Newtonian mechanics A. R French Introductory ll'KHHS o o o o o senes 531.01 FRE OOOOOOO OOOOOOOOOOO OOOOOOOOOOO OOOOOOOOOOO OOOOOOOOOOO OOOOOOOOOOO OOOOOOOOO OOOOOOO I his tcxl prescnts the import m bot h cUlMlol and modern physics as relevani today, atartlng fiom ibout niotion and ending with an problcms in murinml dynamics. i lu- prime empha i the lopment and usc of New. i ii a cornpletc llret-yaar tnechanka couim and contains manj iiiggesdona i<>r further rcadh Aboul tbC Aulhor \ i eived bot h his B. A. Ph. D. degrees from ( amhridge i nivenlty, where he cootinued asa member of the Physics i tepartment from 1948 undi 1955. b ■ic joincd the staff of the l m. : South PhyalGS and became l»*)2 hc wenl to mi i Profc I was appointed t wo years later. has devoted ai half his timc during the past ala lopment of the \t i i . Introdw but his eapertaaca h bj do dm purel . He hai had extc carch assignments with the Cavendlah i aboratoriea, the Manhattan Project, and the U.K. Vtomk Eoergj Reseercb iblishmcnt THE LIBRARY THE HARRIS COLLEGE CORPORATION STREET, PRESTON AH liooks must be Returned to «he College Library or Renewed not later iban the last date sbown below. Re. Re W n< NGV !974 -6. OCT. 19J 975 ! FEB. 197L. -?■ M 197 . 19. m. [991 -3. MAY 1976 ■ 2 -0 ■ 7. I 21. FED. 1997 j _ The M.I.T. Introductory Physics Series Special RelatiVity A.P.FRENCH Vibrations and Waves A.P.FRENCH Newtonian Mechanics A.P.FRENCH TheM.I.T. Introductory Physics Series Newtonian Mechanics A. E French _ PROFESSOR OF MATHEMATICS, THE MASSACHUSETTS INSTITUTE OF TECHNOLOGY Nelson TIIOMAS NELSON AND SONS LTD 36 Park Street London wIy 4de po Box 18123 Nairobi Kenya THOMAS NELSON (AUSTRALIA) LTD 597 Little Collins Street Melbourne 3000 THOMAS NELSON AND SONS (CANADA) LTD 8 1 Curlew Drive Don Mills Ontario THOMAS NELSON (NIGERIA) LTD po Box 336 Apapa Lagos THOMAS NELSON AND SONS (SOUTH AFR1CA) (PROPRIETARY) LTD 51 Commissioner Street Johanncsburg Copyright © 1971, 1965 by the Massachusetts Institute of Technology First published in Great Britain 1971 sbn 17 771074 8 (paper) L^bn 17 761075 1 (boards) PROTON per ■ ic' 47809 — — » 551 we Made and printed by William Clowes & Sons Ltd, Beccles, Colchester and London Contents Preface xi Prologue 3 PART I THE APPROACH TO NEWTONIAN DYNAMICS 1 A universe of particles 21 The particulate view 21 Eleclrons and nucleons 24 Atomic nuclei 25 Atoms 26 Molecules; licing cells 28 Sand and dust 31 Other lerrestrial objecls 32 Planels and satellifes 33 Stars 35 Galaxies 36 PROBLEMS 38 2 Space, time, and motion 43 What is motion? 43 Frames ofreference 46 Coordinale systems 48 Combination ofvector displacements 53 The resolution ofvectors 56 Vector addilion and the properties ofspace 59 Time 61 Uni is and standards oflength and time 63 Space-time graphs 66 Velocily 67 lnslantaneous velocily 68 Relatlve velocily and relative motion 72 Planetary motions: Ptolemy versus Copernicus 74 PROBLEMS 78 3 Accelerated motions 85 Acceleralion 85 The analysis of straight-line motion 87 A comment on exlraneous roots 93 Trujectory problems in two dimensions 95 Free fail of individual aloms 98 Other features of motion in free fail 102 Uniform circular motion 105 Velocily and acceleralion in polar coordinales 106 PROBLEMS 108 4 Forces and equilibrium 115 Forces in static equilibrium 116 Unit s offorce 118 Ekuilibrium conditions; forces as vector s 119 Action and reaction in the contact of objects 123 Rotational ekuilibrium; loraue 124 Forces without contact; weight 128 Pulleys and slrings 130 problems 132 5 The various forces of nature 139 The hasic lypes of forces 139 Gravitatkmal forces 140 Electric and magnetic forces 145 Nuclear forces 147 Forces between neutral atoms 148 Con laci forces 1 50 Frictional contact forces 152 Concluding remarks 154 problems 154 VI 6 Force, inertia, and motion 161 The principle of inertia 161 Force and inertial mass: Newlon's law 164 Some comments on Newlon's law 167 Scales ofmass and force 170 The effect of a continuing force 173 The incariance of Newton' s law; relatioily 173 Incar iance with specific force laws 176 Newlon's law and time reuersal 178 Concluding remarks 180 problems 181 PART II CLASSICAL MECHANICS AT WORK 7 Using Newton's law 187 Some inlroduclory examples 188 Motion in two dimensions 194 Mo I ion in a circle 198 Curoilinear motion with changing speed 200 Circular pai lis of charged particles in uniform magnelic fields 202 Charged particle in a magnelic fielcl 205 Mass spectrographs 206 The fracture ofrapidly rotating objects 208 Motion against resistive forces 210 Deiailed analysis ofresisted motion 213 Motion gocerned by viscosily 218 Growlh and decay ofresisted motion 221 Air resistance and "independence of molions" 225 Simple harmonic motion 226 More about simple harmonic motion 231 problems 234 8 Universal gravitation 245 The discovery of universal gravitation 245 The orbit s of the planet s 246 Planetary periods 249 Kepler's third law 252 The moon and the apple 256 Finding the dislance to the moon 259 The gracitational attraciion of a large sphere 261 Other satellites of the eartli 265 The ualue of G, and the mass of the earth 268 Local uariations of g 270 The mass of the sun 274 Vll Finding the dislance to the sun 275 Mass and weight 279 Weightlessness 285 Learning about other planets 286 The moons ofJupiter 288 The discovery ofNeptune 291 Gravilation outside the solar system 295 Einstein's theory of gravilation 299 PROBLEMS 301 \ 9 Collisions and conservation laws 307 The laws of impact 308 The conservation of linear momentum 309 Momentum as a vector guantity 310 Action, reaction, and impulse 313 Extending the principie of momentum conservation 318 The force exerted hy a stream ofparticles 321 Reaction from a fluid jet 324 Rocket propulsion 327 Collisions and f rames ofreference 331 Kinetic energy in collisions 333 The zero-momentum frame 335 Collision processes in two dimensions 339 Elastic nuclear collisions 342 Inelastic and explosive processes 346 What is a collisionl 351 lnteracting particles subject t o external forces 352 The pressure of a gas 354 The neutrino 356 PROBLtMS 357 10 Energy conservation in dynamics; vibrational motions 367 Introduction 367 lntegrals of motion 368 Work, energy, and power 373 Gravitational potential energy 376 More about one-dimensional situations 379 The energy methodfor one-dimensional motions 381 Some examples of the energy method 384 The harmonic oscillator by the energy method 393 Small oscillations in general 395 The linear oscillator as a two-body problem 397 Collision processes inuolving energy storage 400 The diatomic molecule 405 PROBLEMS 411 Vlll 1 1 Conservative forces and motion in space 423 Extending the concepl of conservative forces 423 Acceleration of two connected masses 425 Object moving in a verlical circle 426 An experimenl by Galileo 429 Mass on a puraboiic track 431 The simple pendulum 434 The pendulum cis a harmonic oscillator 437 The pendulum with larger amplitudo 440 Universal gravitation: a conservative central force 442 A gravitating spherical shell 446 A gravitating sphere 450 Escape velocities 453 More aboul the cri teria for conservative forces 457 Fields 461 Equipotential surfaces and the gradien! of potential energy 463 Motion in conservative fields 466 The effect of dissipative forces 470 Gauss's law 473 Applications of Gauss's theorem 476 PROBLEMS 478 PART III SOME SPECIAL TOPICS 12 Inertial forces and non-inertial frames 493 Motion observed from unaccelerated frames 494 Motion observed front an accelerated frame 495 Accelerated frames and inertial forces 497 Accelerometers 501 Accelerating frames and gravity 504 Centrif ugal force 507 Centrifuges 511 Coriolis forces 514 Dynamics on a merry-go-round 518 General ecjuation of motion in a rotating frame 519 The earth as a rotating reference frame 524 The tides 531 Tidal heights; effect of the sun 535 The search for a fundamental inertial frame 538 Speculations on the origin of inertia 542 PROBLEMS 546 13 Motion under central forces 555 Basic features of the problem 555 The law ofequal areas 557 The conservation of angular momentum 560 IX Energy conseruation in ceniral force molions 563 Vse of the effectiue potential-energy citrues 565 Bounded orbtts 568 Unbcnmded orbits 569 Circular orbtts in an inuerse-square force field 572 Small perturbation of a circular orbit 574 The elliptic orbits ofthe planels 577 Deducing the inuerse-scpiare law from the eltipse 583 Elliptic orbits: analylical treatment 585 Energy in an elliptic orbit 589 Molion near the earth's surface 591 Interplcmetciry transfer orbits 592 Calculaling an orbit from inilial condilions 595 A family of relai ed orbits 596 Central force motion as a two-body problem 598 Deducing the orbit from the force law 600 Rutherford scattering 604 Cross sections for scattering 609 Alpha-particle scattering (Geiger and Marsden, Philosophical Magazine excerpts) 612 An historical note 615 problems 617 14 Extended systems and rotational dynamics 627 Momentum and kinetic energy of a many-particle system 628 Angtilar momentum 632 Angular momentum as a fundamental quanlity 636 Conseruation of angular momentum 639 Moments of inertia of extended objects 643 Two theorems concerning moments of inertia 647 Kinetic energy ofrotaling objects 651 Angular momentum conseruation and kinetic energy 654 Torsional oscillations and rigid pendulums 659 Motion under combined forces and tonjues 664 Impu/sioe forces and torgues 668 Bachground to gyroscopic motion 671 Gyroscope in steady precession 677 More about precessional motion 680 Gyroscopes in nauigation 683 Atoms and nuclei as gyroscopes 686 Gyroscopic motion in teriris of F = »ia 688 Niitation 691 The precession of the equinoxes 694 problems 700 Appendix 709 Bibliography 713 Answers to problems 723 lndex 733 Preface the work of the Education Research Center at M.I.T. (formerly the Science Teaching Center) is concerned with curriculum im- provement, with the process of instruction and aids thereto, and with the learning process itself, primarily with respect to students at the college or university undergraduate level. The Center was established by M.I.T. in 1960, with the late Professor Francis L. Friedman as its Director. Since 1961 the Center has been sup- ported mainly by the National Science Foundation; generous support has also been received from the Kettering Foundation, the Shell Companies Foundation, the Victoria Foundation, the W. T. Grant Foundation, and the Bing Foundation. The M.I.T. Introductory Physics Series, a direct outgrowth of the Center's work, is designed to be a set of short books that, taken collectively, span the main areas of basic physics. The series seeks to emphasize the interaction of experiment and intui- tion in generating physical theories. The books in the series are intended to provide a variety of possible bases for introductory courses, ranging from those which chiefly emphasize classical physics to those which embody a considerable amount of atomic and quantum physics. The various volumes are intended to be compatible in level and style of treatment but are not conceived as a tightly knit package; on the contrary, each book is designed to be reasonably self-contained and usable as an individual com- ponent in many different course structures. XI The text material in the present volume is designed to be a more or less self-contained introduction to Newtonian mechanics, such that a student with little or no grounding in the subject can, by beginning at the beginning, be brought gradually to a level of considerable proficicncy. A rough guide to the possible use of the book is suggested by its division into three parts. Part I, The Approach to Newtonian Dynamics, is intended to serve two pur- poses. First, it does discuss the basic concepts of kinematics and dynamics, more or less from scratch. Second, it seeks to place the study of mechanics squarely in the context of the world of physical phenomena and of neccssarily imperfect physical theories. This is a conscious reaction, on the author's part, against the preserta- tion of mechanics as "applied mathematics," with the divorce- ment from reality and the misleading impression of rigor that this has engendered in generations of students (especially, alas, those brought up in the British educational system). The student who already has some expertise in using Newton's laws will find little of an analytical or quantitative sort to learn from Part I, but he may still derive some value and interest from reading through it for its broader implications. Part II, Classical Mechanics at Work, is undoubtedly the heart of the book. Some instructors will wish to begin here, and relegate Part I to the status of background reading. The initial emphasis is on Newton's second law applied to individual objccts. Later, the emphasis shifts to systems of two or more particlcs, and to the conservation laws for momentum and energy. A fairly lengthy chapter is devoted to the subject that deserves pride of place in the whole Newtonian scheme— the theory of universal gravitation and its successes, which can still be appreciated as a pinnacle in man's attempts to discover order in the vast universe in which he finds himself. Part III, Some Special Topics, concerns itself with the prob- lems of noninertial frames, central-force motions, and rotational dynamics. Most of this material, except perhaps the fundamental features of rotational motion and angular momentum, could be regarded as optional if this book is used as the basis of a genuinely introduetory presentation of mechanics. Undoubtedly the book as a whole contains more material than could in its entirety be covered in a one-term course; one could, however, consider using Parts I and II as a manageablc paekage for beginners, and Parts II and III as a text for students having some prior preparation. XII One of the great satisfactions of classical mechanics lies in the vast range and variety of physical systems to which its princi- ples can be applied. The attempt has been made in this book to make explicit reference to such applications and, as in other books in this series, to "document" the presentation with appropriate citations from original sources. Enriched in this way by its own history, classical mechanics has an excitement that is not, in this author's view, surpassed by any of the more recent fields of physical thory. This book, like the others in the series, owes much to the thoughts, criticisms, and suggestions of many people, both students and instructors. A special acknowledgment in connection with the present volume is due to Prof. A. M. Hudson, of Occiden- tal College, Los Angeles, who worked with the present author in the preparation of the preliminary text from which, five years later, this final version evolved. Grateful thanks are also due to Eva M. Hakala and William H. Ingham for their invaluable help in preparing the manuscript for publication. A. P. FRENCH Cambridge, Massachusetts July 1970 XIII Newtonian mechanics In the Beginning was Mechanics. max von laue, History ofPhysics (1950) / offer this work as the mathematical principles of philosophy, for the whole burden of philosophy seems to consist in this — from the phenomena ofmotions to investigate theforces of nature, and thenfrom these forces to demonstrate the other phenomena. newton, Preface to the Principia (1686) Prologue one of the most prominent features of the universe is motion. Galaxies have motions with respect to other galaxies, all stars have motions, the planets have distinctive motions against the background of the stars, the events that capture our attention most quickly in everyday life are those involving motion, and even the apparently inert book that you are now reading is made up of atoms in rapid motion about their equilibrium positions. "Give me matter and motion," said the seventeenth-century French philosopher Rene Descartes, "and I will construct the universe." There can be no doubt that motion is a phenomenon we must learn to deal with at all levels if we are to understand the world around us. . Isaac Newton developed a precise and powerful theory j regarding motion, according to which the changes of_motion of any object are the result of J orces acting on itfln so doing he created the subject with "wmcrTthis book is coBeerned and which is called classical or Newtonian mechanics. I It was a landmark in the history of science, because it replacea a merely descriptive account of phenomena with a rational and marvelously successful scheme of cause and effect. Indeed, the strict causal nature of Newtonian mechanics had an impressive influence in the develop- ment of Western thought and civilization generally, provoking fundamental questions about the interrelationships of science, philosophy, and religion, with repercussions in social ideas and other areas of human endeavor. Classical mechanics is a subject with a fascinating dual character. For it starts out from the kinds of everyday experiences that are as old as mankind, yet it brings us face to face with some of the most profound questions about the universe in which we find ourselves. Is it not remarkable that the fiight of a thrown pebble, or the fail of an apple, should contain the clue to the mechanics of the heavens and should ultimately involve some of the most basic questions that we are able to formulate about the nature of space and time? Sometimes mechanics is presented as though it consisted merely of the routine application of self- evident or revealed truths. Nothing could be further from the case; it is a superb example of a physical theory, slowly evolved and refined through the continuing interplay between observation and hypothesis. The richness of our first-hand acquaintance with mechanics is impressive, and through the partnership of mind and eye and hand we solve, by direct action, innumerable dynamical problems without benefit of mathematical analysis. Like Moliere's famous character, M. Jourdain, who learned that he had been speaking prose all his life without realizing it, every human being is an expert in the consequences of the laws of mechanics, whether or not he has ever seen these laws written down. The skilled sports- man or athlete has an almost incredible degree of judgment and control of the amount and direction of muscular effort needed to achieve a desired result. It has been estimated, for example, that the World Series baseball championship would have changed hands in 1962 if one crucial swing at the ball had been a mere millimeter lower. ' But experiencing and controlling the motions of objects in this very personal sense is a far cry from analyzing them in terms of physical laws and equations. It is the task of classical mechanics to discover and formulate the essential principles, so that they can be applied to any situation, par- ticularly to inanimate objects interacting with one another. Our intimate familiarity with our own muscular actions and their consequences, although it represents a kind of understanding (and an important kind, too), does not help us much here. The greatest triumph of classical mechanics was Newton's own success in analyzing the workings of the solar system — a feat immortalized in the famous couplet of his contemporary and admirer, the poet Alexander Pope : ■P. Kirkpatrick, Am. J. Phys., 31, 606 (1963). 4 Prologue Nature and Nature's Laws Iay hid in night God said "Let Newton be," and all was light. 1 Men had observed the motions of the heavenly bodies since time immemorial. They had noticed various regularities and had learned to predict such things as conjunctions of the planets and eclipses of the sun and moon. Then, in the sixteenth century, the Danish astronomer Tycho Brahe amassed meticulous records, of unprecedented accuracy, of the planetary motions. His assistant, Johannes Kepler, after wrestling with this enormous body of in- formation for years, found that all the observations could be summarized as follows: 1 . The planets move in ellipses having the sun at one focus. 2. The line joining the sun to a given planet sweeps out equal areas in equal times. 3. The square of a planet's year, divided by the cube of its mean distance from the sun, is the same for all planets. This represented a magnificent advance in man's knowledge of the mechanics of the heavens, but it was still a description rather than a theory. Why? was the question that still looked for an answer. Then came Newton, with his concept of force as the cause of changes of motion, and with his postulate of a particular law of force — the inverse-square law of gravitation. Using these he demonstrated how Kepler's laws were just one consequence of a scheme of things that also included the falling apple and other terrestrial motions. (Later in this book we shall go into the de- tails of this great achievement of Newton's.) I f universal gravitation had done no more than to relate planetary periods and distances, it would still have been a splen- did theory. But, like any other good theory in physics, it had predictive value; that is, it could be applied to situations besides the ones from which it was deduced. Investigating the predic- tions of a theory may involve looking for hitherto unsuspected phenomena, or it may involve recognizing that an already familiar phenomenon must fit into the new framework. In either case the theory is subjected to searching tests, by which it must 'To which there is the almost equally famous, although facetious, riposte: It did not last; the Devil, howling "Ho, Let Einstein be!" restored the status quo. (Sir John Squire) 5 Prologue stand or fail. With Newton's theory of gravitation, the initial tests resided almost entirely in the analysis of known effects— but what a list! Here are some of the phenomena for which Newton proceeded to give quantitative explanations: 1. The bulging of the earth and Jupiter because of their rotation. 2. The variation of the acceleration of gravity with latitude over the earth's surface. 3. The generation of the tides by the combined action of sun and moon. 4. The paths of the comets through the solar system. 5. The slow steady change in direction of the earth's axis of rotation produced by gravitational torques from the sun and moon. (A complete cycle of this variation takes about 25,000 years, and the so-called "precession of the equinoxes" is a mani- festation of it.) This marvelous illumination of the workings of nature represented the last part of Newton's program, as he described it in our opening quotation ". . . and then from these forces to demonstrate the other phenomena." This modest phrase con- ceals not only the immensity of the achievement but also the magnitude of the role played by mathematics in this develop- ment. Newton had, in the theory of universal gravitation, created what would be called today a mathematical model of the solar system. And having once made the model, he followed out a host of its other implications. The working out was purely mathematical, but the final step— the test of the conclusions— involved a return to the world of physical experience, in the detailed checking of his predictions against the quantitative data of astronomy. Although Newton's mechanical picture of the universe was amply confirmed in his own time, he did not live to see some of its greatest triumphs. Perhaps the most impressive of these was the use of his laws to identify previously unrecognized mem- bers of the solar system. By a painstaking and lengthy analysis of the motions of the known planets, it was inferred that dis- turbing influences due to other planets must be at work. Thus it was that Neptune was discovered in 1846, and Pluto in 1930. In each case it was a matter of deducing where a telescope should be pointed to reveal a new planet, identifiable through its chang- ing position with respect to the general background of the stars. 6 Prologuc What more striking and convincing evidence could there be that the theory works? Probably everyone who reads this book has some prior acquaintance with classical mechanics and with its expression in mathematically precise statements. And this may make it hard to realize that, as with any other physical theory, its development was not just a matter of mathematical logic applied undis- criminatingly to a mass of data. Was Newton inexorably driven to the inverse-square Iaw? By no means. It was the result of guesswork, intuition, and imagination. In Newton's own words: "I began to think of gravity extending to the orbit of the Moon, and . . . fromKepler's Ruleof the periodictimesofthePlanets . . . I deduced that the forces which keep the Planets in their orbits must be reciprocally as the squares of their distances from the centers about which they revolve; and thereby compared the force requisite to keep the Moon in her orbit with the force of gravity at the surface of the Earth, and found them to answer pretty nearly." An intellectual leap of this sort — although seldom as great as Newton's — is involved in the creation of any theory or model. It is a process of induction, and it goes beyond the facts immediately at hand. Some facts may even be temporarily brushed aside or ignored in the interests of pursuing the main idea, for a partially correct theory is often better than no theory at all. And at all stages there is a constant interplay between experiment and theory, in the process of which fresh observations are continually suggesting themselves and modification of the theory is an ever-present possibility. The following diagram, the relevance of which goes beyond the realm of classical mechanics, suggests this pattern of man's investigation of matter and motion. Laws of Motion — INDUCTION - Observations and Experiments 1 of J Laws of Force •DEDUCTION- Predictions Mathematical Models I The enormous success of classical mechanics made it seem, at one stage, that nothing more was needed to account for the whole world of physical phenomena. This belief reached a pinnacle toward the end of the nineteenth century, when some 7 Prologue optimistic physicists felt that physics was, in principle, complete. They could hardly have chosen a more unfortunate time at which to form such a conclusion, for within the next few decades physics underwent its greatest upheaval since Newton. The dis- covery of radioactivity, of the electron and the nucleus, and the subtleties of electromagnetism, called for fundamentally new ideas. Thus we know today that Newtonian mechanics, like every physical theory, has its fundamental limitations. The analysis of motions at extremely high speeds requires the use of modified descriptions of space and time, as spelled out by Albert Einstein's special theory of relativity. In the analysis of phe- nomena on the atomic or subatomic scale, the still more drastic modifications described by quantum theory are required. And Newton's particular version of gravitational theory, for all its success, has had to admit modifications embodied in Einstein's general theory of relativity. But this does not alter the fact that, in an enormous range and variety of situations, Newtonian mechanics provides us with the means to analyze and predict the motions of physical objects, from electrons to galaxies. Its range of validity, and its limits, are indicated very qualitatively in the figure below. In developing the subject of classical mechanics in this book, we shall try to indicate how the horizons of its application to the physical world, and the horizons of one's own view, can be gradually broadened. Mechanics, as we shall try to present it, is not at all a cut-and-dried subject that would justify its description Cosmological Physics 10-'° m Atom 10 !0 m Galaxy Size 8 Proloeuc as "applied mathematics," in which the rules of the game are given at the outset and in which one's only concern is with apply- ing the rules to a variety of situations. We wish to offer a dif- ferent approach, in which at every stage one can be conscious of working with partial or limited data and of making use of as- sumptions that cannot be rigorously justifled. But this is the essence of doing physics. Newton himself said as much. At the beginning of Book III of the Principia he propounds four "Rules of Reasoning in Philosophy," of which the last runs as follows: "In experimental philosophy we are to look upon proposi- tions inferred by general induction from phenomena as accurately or very nearly true, notwithstanding any contrary hypotheses that may be imagined, till such time as other phenomena occur, by which they may either be made more accurate, or liable to exceptions." The person who waits for complete information is on the way to dooming himself never to finish an experiment or to construct a useful theory. Lest this should be taken, however, as an encouragement to slipshod or superficial thinking, we shall end this introduction with a little fable due to George Polya. 1 He writes as a mathematician, but the moral for physicists (and others) is clear. The Logician, the Mathematician, the Physicist, and the Engineer "Look at this mathematician," said the logician. "He observes that the first 99 numbers are less than 100 and infers, hence, by what he calls induction, that all numbers are less than a hundred." "A physicist believes," said the mathematician, "that 60 is divisible by 1, 2, 3, 4, 5, and 6. He examines a few more cases, such as 10, 20, and 30, taken at random (as he says). Since 60 is also divisible by these, he considers the experimental evidence sufficient." "Yes, but look at the engineers," said the physicist. "An engineer suspected that all odd numbers are prime numbers. At any rate, 1 can be considered as a prime number, he argued. Then there come 3, 5, and 7, all indubitably primes. Then there comes 9; an awkward case; it does not seem to be a prime num- 'This cautionary tale is to be found in a book entitled Induction and Analogy in Mathematics, Princeton University Press, Princeton, N.J., 1954. This volume and its companion, Patterns of Plausible Inference, make delightful reading for any scientist. 9 Prologuc ber. Yet 1 1 and 13 are certainly primes. 'Corning back to 9,' he said, 'I conclude that 9 must be an experimental error.' " But having done his teasing, Polya adds these remarks. It is only too obvious that induction can lead to error. Yet it is remarkable that induction sometimes leads to truth, since the chances of error seem so overwhelming. Should we begin with the study of the obvious cases in which induction fails, or with the study of those remarkable cases in which induction succeeds? The study of precious stones is understandably more attractive than that of ordinary pebbles and, moreover, it was much more the precious stones than the pebbles that led the mineralogists to the wonderful science of crystallography. With that encouragement, we shall, in Chapter l, begin our approach to the study of classical mechanics, which is one of the most perfect and polished gems in the physicist's treasury. We end this Prologue, however, with some preparatory exercises. EXERCISES-HORS D'OEUVRES The literal meaning of the phrase "hors d'oeuvre" is "outside the work." The exercises below correspond exactly to that definition, although it is hoped that they will also whet the appetite as hors d'oeuvres should. They deal mostly with order- of-magnitude estimates (i.e., estimates to the nearest power of 10) and judicious approximations— things that play an important role in a physicist's approach to problems but seldom get em- phasized or systematically presented in textbooks. For example, everybody learns the binomial theorem, but how many students think of it as a useful tool for obtaining a quite good value for the hypotenuse of a right triangle, by the approximation ^ + b^'^a{\+^ where we assume b < a? (Even in the worst possible case, with b = a, the result is wrong by only about 6 percent — 1.5 instead of 1.414 . . . .) Moreover, it takes practice and some conscious effort to develop the habit of assessing, quite crudely, the magni- tudes of quantities and the relative importance of various pos- sible effects in a physical system. For example, in dealing with objects moving through liquids, can one quickly decide whether 10 Prologue viscosity or turbulence is going to be the chief source of resistance for an object of given speed and linear dimensions? An awareness of the effects of changes of scale can give valuable insights into the properties of systems. [A beautiful example of this is the well-known essay by J. B. S. Haldane, "On Being the Right Size," which is reprinted in The World of Mathematics, Vol. II (J. R. Newman, ed.), Simon and Schuster, New York, 1956.] By the use of such methods and ways of thought one can deepen one's appreciation of physical phenomena and can improve one's feeling for what the world is like and how it behaves. It is surprising how much one can do with the help of a relatively small stock of primary information — which might in- clude such items as the following: Physical Magnitudes Gravitational acceleration (g) Densities of solids and liquids Density of air at sea level Length of day Length of year Earth's radius Angle subtended by finger thickness at arm's length Thickness of paper Mass of a paperclip Highest mountains, deepest oceans Earth-moon separation Earth-sun separation Atmospheric pressure Avogadro's number Atomic masses Linear dimensions of atoms Molecules/cm 3 i n gas at STP Atoms/cm 3 in solids Elementary charge (e) Electron mass Speed of light Wavelength of light 10 m/sec' 2 10 3 -10 4 kg/m 3 1 kg/m 3 (approx.) 10 5 sec (approx.) 3.16 X 10 7 sec « I0 75 sec 6400 km 1° (approx.) 0.1 mm (approx.) 0.5 g (approx.) 10 km (approx.) 3.8 X 10 5 km 1.5 X 10 8 km Equivalent to weight of 1 kg/cm 2 or a 10-m column of water 6.0 X 10 23 1.6 X 10- 27 kgto 4 X 10 _25 kg 10 _,0 m (approx.) 2.7 X 10 13 10 23 (approx.) 1.6 X lO -1 9 C 10~ 30 kg (approx.) 3 X 10 8 m/sec 6 X 10 -7 m (approx.) 11 EKcrcises- hors d'oeuvres Malhematical Magnitudes w 2 m 10 log,,, 4 m 0.60 e « 2.7 log, o e ~ 0.43 log 10 2«0.30 log, T«0.50 log, o 3 m 0.48 log c 10 « 2.3 Angle (radians) = arc length/radius. Full circle = 2jrrad. 1 rad « 0.16 X full circle « 57°. Solid angle (steradians) = area/(radius) 2 . Full sphere = 4*- sr. 1 sr ~ 0.08 X full sphere. Approximations Binomial theorem: Forx« 1, (1 + x) n m 1 + «* e.g., (1 + x) 3 « 1 + 3* (1 -x)" 2 a 1 - **« (1 +x)- 1/2 For b « a, (a + b)" - fl*(l + jjjY ~ a n (l+n?\ Othcr expansions: e 3 For 6 « 1 rad, sin fl w > 6 e 2 cos d « 1 ► 1 2 For a: « 1, log« (1 + x) ~ x log, (l +Jc)« 0.43x No answers are given to the problems that follow. For most of them, you yourself will be the best judge. You may want to turn to an encyclopedia or other reference book to check some of your assumptions or conclusions. If you are not prepared at this point to tackle them all, don't worry; you can always return to them later. / What is the order of magnitude of the number of times that the earth has rotated on its axis since the solar system was formed? 2 During the average lifetime of a human being, how many heart- beats are there ? How many breaths ? 3 Make reasoned estimates of (a) the total number of ancestors ytou would have (ignoring inbreeding) since the beginning of the human race, and (b) the number of hairs on your head. 4 The present world population (human) is about 3 X 10°. (a) How many squarc kilometers of land are there per person? 12 Prologue How many feet long is the side of a square of that area? (b) If one assumes that the population has been doubling every 50 years throughout the existence of the human race, when did Adam and Eve start it all ? If the doubling every 50 years were to continue, how long would it be before people were standing shoulder to shoulder over all the land area of the world? 5 Estimate the order of magnitude of the mass of (a) a speck of dust ; (b) a grain of sak (or sugar, or sand); (c) a mouse; (d) an elephant; (e) the water corresponding to 1 in. of rainfall over 1 square mile; (0 a small hill, 500 ft high; and (g) Mount Everest. 6 Estimate the order of magnilude of the number of atoms in (a) a pin's head, (b) a human being, (c) the earth's atmosphere, and (d) the whole earth. 7 Estimate the fraction of the total mass of the earth that is now in the form of living things. 8 Estimate (a) the total volume of ocean water on the earth, and (b) the total mass of sah in all the oceans. 9 It is estimated that there are about 10 80 protons in the (known) universe. If all these were lumped into a sphere so that they were just touching, what would the radius of the sphere be? Ignore the spaces left when spherical objects are packed and takc the radius of a proton to be about 10~ 15 m. 10 The sun is losing mass (in the form of radiant energy) at the rate of about 4 million tons per second. What fraction of its mass has it lost during the lifetime of the solar system ? 11 Estimate the time in minutcs that it would take for a theatre audience of about 1000 people to use up 10% of the available oxygen if the build- ing were sealed. The average adult absorbs about one sixth of the oxygen that he or she inhales at each breath. 12 Solar energy falls on the earth at the rate of about 2 cal/cm 2 /min. Estimate the amount of power, in megawatts or horsepower, repre- sented by the solar energy falling on an area of 100 square miles — about the area of a good-sized city. How would this comparc with the total power requirementsofsuch a city? (1 cal = 4.2 J; 1 W = U/sec; 1 hp = 746 W.) 13 Starting from an estimate of the total mileage that an automobile tire will give before wearing out, estimate what thickness of rubber is worn off during one revolution of the wheel. Consider the possible physical significance of the result. (With acknowledgment to E. M. Rogers, Physics for the lnquiring Mind, Princeton University Press, Princeton, N. J., 1%0.) 14 An ine.vpensive wristwatch is found to lose 2 min/day. (a) What is its fractional dcviation from the correct rate? 13 Exercises — hors d'oei v re s (b) By how much could the length of a ruler (nominally 1 ft long) differ from exactly 12 in. and still be fractionally as accurate as the watch ? 15 The astronomer Tycho Brahe made observations on the angular posilions of stars and planets by using a quadrant, with one peephole at its center of curvature and another peephole mounted on the arc. One such quadrant had a radius of about 2 m, and Tycho's measure- ments could usually be trusted to 1 minute of arc (^g°). What diameter of peepholes would havc been needed for him to attain this accuracy ? 16 Jupiter has a mass about 300 times that of the earth, but its mean density is only about one fifth that of the earth. (a) What radius would a planet of Jupiter's mass and earth's density have? (b) What radius would a planet of earth's mass and Jupiter's density have ? 17 Identical spheres of material are tightly packed in a given volume of space. (a) Consider why one does not need to know the radius of the spheres, but only the density of the material, in order to calculate the total mass contained in the volume, provided that the linear dimensions of the volume are large compared to the radius of the individual spheres. (b) Consider the possibility of packing more material if two sizes of spheres may be chosen and used. (c) Show that the total surface area of the spheres of part (a) does depend on the radius of the spheres (an important consideration in the design of such things as filters, which absorb in proportion to the total exposed surface area within a given volume). 18 Calculate the ratio of surface area to volume for (a) a sphere of radius r, (b) a cube of edge a, and (c) a right circular cylinder of diameter and height both equal to d. For a given value of the volume, which of these shapes has the greatest surface area ? The least surface area? 19 How many seconds of arc does the diameter of the earth subtend at the sun? At what distance from an observer should a football be placed to subtend an equal angle? 20 From the time the lower limb of the sun touches the horizon it takes approximately 2 min for the sun to disappear beneath the horizon. (a) Approximately what angle (exprcssed both in degrees and in radians) does the diameter of the sun subtend at the earth ? (b) At what distance from your eye does a coin of about ^-in. diameter (e.g., a dime or a nickel) just block out the disk of the sun? (c) What solid angle (in steradians) does the sun subtend at the earth? 14 Prologue 21 How many inches per mile does a terrestrial great circle (e.g., a meridian of longitude) deviate from a straight line ? 22 A crude measure of the roughness of a nearly spherical surface could be defined by Ar/r, where Ar is the height or depth of local irregularities. Estimate this ratio for an orange, a ping-pong ball, and the earth. 23 What is the probability (expressed as 1 chance in 10") that a good- sized meteorite falling to earth would strike a man-made structure? A human ? 24 Two students want to measure the speed of sound by the following procedure. One of them, positioned some distance away from the other, sets off a firecracker. The second student starts a stopwatch when he sees the flash and stops it when he hears the bang. The speed of sound in air is roughly 300 m/scc, and the students must admit the possibility of an error (of undetermined sign) of perhaps 0.3 sec in the elapsed time recorded. If they wish to keep the error in the measured speed of sound to within 5%, what is the minimum distance over which they can perform the experiment? 25 A right triangle has sides of length 5 m and 1 m adjoining the right angle. Calculate the length of the hypotenuse from the binomial ex- pansion to two terms only, and estimate the fractional error in this approximate result. 26 The radius of a sphere is measured with an uncertainty of 1%. What is the percentage uncertainty in the volume? 27 Construct a piece of semilogarithmic graph paper by using the graduations on your slide rulc to mark off the ordinates and a normal ruler to mark off the abscissa. On this piece of paper draw a graph of the function y = 2 X . 28 The subjective sensations of loudness or brightness have bcen judged to be approximately proportional to the logarithm of the intensity, so that equal mulliples of intensity are associated with equal arithmetic increases in sensation. (For example, intensities propor- tional to 2, 4, 8, and 16 would correspond to equal increases in sensa- tion.) In acoustics, this has led to the measurement of sound intensities in decibels. Taking as a reference value the intensity /o of the faintest audible sound, the decibel level of a sound of intensity / is defined by the equation dB = lOlogio © (a) An intolerable noise level is represented by about 120 dB. By what factor does the intensity of such a sound exceed the threshold intensity /o? 15 Exercises — hors d'oeuvres (b) A similar logarithmic scale is used to describe the relative brightness of stars (as seen from the earth) in terms of magnitudes. Stars differing by "one magnitude" have a ratio of apparent brightness equal to about 2.5. Thus a "first-magnitude" (very bright) star is 2.5 times brighter than a second-magnitude star, (2.5) 2 times brighter than a third-magnitude star, and so on. (These differences are due largely to differences of distance.) The faintest stars detectable with the 200-in. Palomar telescope are of about the twenty-fourth magnitude. By what factor is the amount of light reaching us from such a star less than we receive from a first-magnitude star ? 29 The universe appears to be undergoing a general expansion in which the galaxies are receding from us at speeds proportional to their dis- tances. This is described by Hubble's law, v = «r, where the con- stant a corresponds to v becoming equal to the speed of light, c (= 3 X 10 8 m/sec), at r « 10 26 m. This would imply that the mean mass per unit volume in the universe is decreasing with time. (a) Suppose that the universe is represented by a sphere of volume V at any instant. Show that the fractional increase of volume per unit time is given by 1 dV = 3a V dt (b) Calculate the fractional decrease of mean density per second and per century. 30 The table lists the mean orbit radii of successivc planets expressed in terms of the earth's orbit radius. The planets are numbered in order («): Planet r/rg 1 Mercury 0.39 2 Venus 0.72 3 Earth 1.00 4 Mars 1.52 5 Jupiter 5.20 6 Saturn 9.54 7 Uranus 19.2 (a) Make a graph in which \og(r/r E ) is ordinate and the number n is abscissa. (Or, alternatively, plot values of r /r e against n on semi- logarithmic paper.) On this samc graph, replot the points for Jupiter, Saturn, and Uranus at values of n increased by unity (i.e., at n = 6, 7, and 8). The points representing the seven planets can thcn be reason- ably well fitted by a straight line. (b) If n = 5 in the revised plot is taken to represent the asteroid belt between the orbits of Mars and Jupiter, what value of r/r E would 16 Prologue your graph imply for this ? Compare with the actual mean radius of the asteroid belt. (c) If n = 9 is taken to suggest an orbit radius for the next planet (Neptune) beyond Uranus, what value of r/re would your graph imply ? Compare with the observed value. (d) Consider whether, in the light of (b) and (c), your graph can be regarded as the expression of a physical law with predictive value. (As a matter of history, it was so used. See the account of the dis- covery of Neptune near the end of Chapter 8.) PHILOSOPHLE NATURALIS PRINCIPI A MATHEMATICA Autore J S. NEWTON, Irin. Coli. Cantab. Soc. Mathefeos Profeflbre Lucafiano, & Socictatis Regalis Sodalt. IMPRIMATUR S. P E P Y S, Reg. Soc. P R JE S E S. Juiii 5. 1686. L N D I N I, Juflii Soaetatis Rcgi* ac Typis Jofepbi Strealcr. Proftat apud plures Bibliopolas. Amo MDCLXXXVII. Facsimile oflhe title page oflhe firsl edilion of Newion's Principia (published 1687). It may be seen thal Ihe work was officially accepted by Ihe Royal Sociely of London in July, 1686, when ils president was thefamous diarist Samuel Pepys (who was also Secretary to Ihe Admirally at Ihe time). PartI The approach to Newtonian dynamics // seems probable to me, that God in the Beginningfortrid Matter in solid, massy, hard, impenetrable, moveable Particles .... newton, Opticks (1730) 1 A universe of particles THE PARTICULATE VI EW the essence of the Newtonian approach to mechanics is that the motion of a given object is analyzed in terms of the forces to which it is subjected by its environment. Thus from the very outset we are concerned with discrete objects of various kinds. A special interest attaches to objects that can be treated as if they are point masses; such objects are called particles. 1 In the strictest sense there is nothing in nature that fits this definition. Nevertheless, you have lived for years in a world of particles — electrons, atoms, baseballs, earth satellites, stars, galaxies — and have an excellent idea of what a particle is. If you have read George Orwell's famous political satire Animal Farm, you may remember the cynical proclamation : "Ali animals are equal, but some animals are more equal than others." In somewhat the same way, you may feel that some particles (electrons or protons, for example) are more particulate than others. But in any case the judgement as to whether something is a particle can only be made in terms of specific questions — specific kinds of experiments and observations. And the answer to the question "Is such and such an object a particle?" is not a clear-cut yes or no, but "It depends." For example, atoms and atomic nuclei will look (i.e., behave) like 'Actually, Newton himself reserved the word "particle" to denote what we might now call "fundamental particles"— atoms and other such natural building blocks — but the usage has since changed. 21 Fig. 1-1 Photograph of a portion of the night sky. (Photograph from the Hale Obsercatories.) particles if you don't hit them too hard. Planets and stars will look like particles (both visually and in behavior) if you get far enough away from them (see Fig. 1-1). But every one of these objects has spatial extension and an internal structure, and there will always be circumstances in which these features must be taken into account. Very often this will be done by picturing a given object not as a single point particle but as an assemblage of such ideal particles, more or less firmly connected to one another. (If the connections are sufficiently strong, it may be possible to make use of another fiction — the ideal "rigid body" — that further simplifies the analysis of rotational motions, in particular.) For the moment, however, we shall restrict ourselves 22 A universe of particles to a consideration of objects that exist as recognizable, individual entities and behave, in appropriate circumstances, as particles in the idealized dynamical sense. What sort of information do we need to build up a good description of a particle? Here are a few obvious items, which we write down without any suggestion that the list is exhaustive (or, for that matter, sharply categorized): 1. Mass 2. Size 3. Shape 4. Internal structure 5. Electric charge 6. Magnetic properties 7. Interaction with other particles of the same kind 8. Interaction with different sorts of particles Partial though that list may be, it is already formidable, and it would not be realistic to tackle it all at once. So we ask a more modest question : What is the smalkst number of properties that suffkes to characterize a particle? If we are concerned with the so-called "elementary" particles (electrons, mesons, etc), the state of charge (positive, negative, or neutral) is an important datum, along with the mass, and these two may be sufficient to identify such a particle in many circumstances. Most other objects, composed of large numbers of atoms, are normally electrically neutral, and in any event the mass alone is for many purposes the only property that counts in considering a particle's dynamic behavior — provided we take the forces acting on it as being independently specified. ' It is, however, useful to know, at least approximately, the size also. Not only is this one of our most informative pieces of data concerning any object, but its magnitude will help to tell us whether, in given circumstances, the particle may reasonably be treated as a point mass. Recognizing, then, that many of the finer details will have to be filled in later, we shall begin with a minimal description in which particles are objects possessing mass and size. Our survey is not exhaustive or detailed. On the contrary, we have sought 'Of course, if we want to treat the forces as being derived from characteristic interactions of the body with its surroundings (e.g., by gravitatton), then the laws of interaction must also be known. That is the subject of Chapter 5. 23 The particulate view to reduce it to a minimum, consistent with illustrating the gen- eral scheme of things, by considering only the masses and the linear dimensions of some typical particles. We shall begin with the smallest and least massive particles and go up the scale until we reach what appears to be a fundamental limit. You will appreciate that this account, brief though it is, draws upon the results of a tremendous amount of painstaking observation and research in diverse fields. A note on units In this book we shall most frequently employ the meter-kilogram- second (MKS) metric system. You are probably already familiar with it, at least for the basic measures of mass, length, and time. If not, you should learn it at this time. We shall, however, make occasional use of other measures. In mechanics the conversion from one system of measurement to another presents no problem, because it is just a matter of applying simple numerical factors. (This is in contrast to electromagnetism, where the particular choice of primary quantities affects the detailed formulation of the theory.) A tabulation of MKS and other units is given in the Appendix. ELECTRONS AND NUCLEONS The principal building blocks of matter from the standpoint of physics and chemistry are electrons, protons, and neutrons. Protons and neutrons are virtually equivalent as constituents of atomic nuclei and are lumped together under the generic title nucleons. The vast amount of research on the so-called ele- mentary particles, and on the structure of nucleons, has not brought forth any evidence for particles notably smaller (or notably less massive) than those that were known to science 50 years ago. Thus, although the study of subatomic particles is a field of very great richness and complexity, filled with bizarre and previously unsuspected phenomena, the microscopic limits of the physical world are still well represented by such familiar particles as electrons and protons. Theelectron,withamassofaboutl(r 30 kg(9.1 X l(T 31 kg to be more precise), is by far the lightest (by more than three 24 A universc of particles powers of 10) of the familiar constituents of matter. (The elusive neutrino, emitted in radioactive beta decay, appears to have no mass at all. This puts it in a rather special category!) The size of the electron is not something that can be unequivocally stated. Indeed, the concept of size is not sharply or uniquely defined for any object. If, however, we regard the electron as a sphere of electric charge, its radius can be estimated to be of the order of 10 -15 m. In our present state of knowledge, the electron can properly be regarded as a fundamental particle, in the sense that there is no evidence that it can be analyzed into other constituents. The nucleon, with a mass of 1.67 X 10~ 27 kg, is the other basic ingredient of atoms. In its electrically charged form — the proton — it is (like the electron) completely stable; that is, it survives indefinitely in isolation. In its electrically neutral form — the neutron — it cannot survive in isolation but decays radio- actively (with a half-life of about 13 min) into a proton, an electron, and a neutrino. The fact that neutrons spontaneously give birth to the constituents of hydrogen atoms has led some cosmologists to suggest that neutrons represent the true primeval particles of the universe — but that is just a speculation. Nucleons have a diameter of about 3 X 10 -15 m — by which we mean that the nuclear matter appears to be confined within a moderately well defined region of this size. Unlike electrons, nucleons seem to have a quite complex internal structure, in which various types of mesons are incorporated. But from the standpoint of atomic physics they can be regarded as primary particles. ATOMIC NUCLEI The combination of protons and neutrons to form nuclei pro- vides the basis for the various forms of stable, ordinary matter as we know it. The smallest and lightest nucleus is of course the individual proton. The heaviest naturally occurring nucleus (that of 238 U)— contains 238 nucleons and has a mass of 4.0 X 10 -25 kg. All nuclei have about the same mass per unit volume, so that their diameters are roughly proportional to the cube roots of the numbers of the nucleons. Thus nuclear diameters cover a range from about 3 X 10~ 15 to 2 X 10~ 14 m. A unit of distance has been defined that is very convenient when dealing with nuclear dimensions. It is named after the 25 Alomic nuclei ATOMS Italian physicist Enrico Fermi 1 : lfermi(F) = H)" 18 m = 10- 13 cm Thus the range of nuciear diameters is from about 3 to 20 F. The density of nuciear matter is enormous. Given that the uranium nucleus has a mass of about 4 X 10 -25 kg and a radius of about 10 F, you can deduce (do it!) that its density is about 10 17 kg/m 3 . This is so vast (it is larger, by a factor of 10 14 , than the density of water) that we really cannot apprehend it, although we now have evidence that some astronomical objects (neutron stars) are composed of this nuciear matter in bulk. A great deal was learned about atomic masses long before it was possible to count individual atoms. From the concepts of valence and chemical combinations, chemists established a relative mass scale based on assigning to hydrogen a mass of 1 . The mole was introduced as that amount of any element or compound whose mass in grams was equal numerically to its relative mass on this scale. Furthermore, from the relative proportions of elements that combined to form compounds, it was known that a mole of any substance must contain the same unique number of atoms (or molecules in the case of compounds) — the number known as Avogadro's constant. But this number was itself unknown. Obviously, if the number could be determined, the mass of an individual atom could be found. The existence of characteristic mass transfers in electrolysis gave corroborative evidence on relative atomic masses but also pointed the way to absolute mass determinations, for it seemed clear that the electrolytic phenomena stemmed from a charac- teristic atomic charge unit. Ali that was necessary was to establish the size of this unit (e)— a feat finally achieved in Millikan's precision measurements in 1909. Some representative atomic mass values are listed in Table 1-1. 'E. Fermi (1901-1954) was the greatest Italian physicist since Galileo and one of the most distinguished scientists of the twentieth century, gifted in both theoretical and experimental work. He achieved popular fame as the man who produced the first self-sustained nuciear chain reaction, at the Uni- versity of Chicago in 1942. 26 A universe of particles TABLE 1-1: ATOMIC MASSES Electrotytic Atomic mass transfer, Charge Approximate mass. Element kg/C per ion relative mass kg H 1.04 X lO" 8 e 1 1.67 X lO" 27 C 6.22 X 10- 8 2e 12 2.00 X 10" 26 O 8.29 X 10- 8 2e 16 2.66 X lO" 26 Na 2.38 X 10~ 7 e 23 3.81 X K)- 26 Al 9.32 X lO" 8 3e 27 4.48 X 10 -26 K 4.05 X 10" 7 e 39 6.49 X lO -26 Zn 3.39 X 10- 7 2e 65 1.09 X 10- 2S Ag 1.118 X 10- 6 e 107 1.79 X 10~ 2S Modern precision measurements of atomic masses are based on mass spectroscopy (see p. 206 for an account of the principles) and are quoted in terms of an atomic mass unit (amu). This is now defined as tV of the mass of the isotope carbon 12. lamu = 1.66043 X 10" 27 kg Since almost all the mass of any atom is concentrated in its nucleus (99.95% for hydrogen, rising to 99.98% for uranium), we can say that to a first approximation the mass of an atom is just the mass of its nucleus. But, in terms of size, the atom represents a leap of many orders of magnitude. Nuclear diam- eters, as we have just seen, are of the order of 10 -14 m. Atomic diameters are typically about 10 4 times larger than this — i.e., of the order of 10 -1 ° m. One way of getting a feeling for what this factor means is to consider that if the dot on a printed letter i on this page is taken to represent a medium-weight nucleus, the outer boundary of the atom is about 10 ft away. Think of a grain of fine sand suspended in the niiddle of your bedroom or study, and you will get a feeling for what that means in three dimensions. (Nuclei are really very small.) It is very convenient to take 10 -10 m as a unit of distance in describing atomic sizes or interatomic distances in solids and other condensed states in which the atoms are closely packed. The unit is named after the nineteenth-century Swedish physicist, A. J. Angstrom: 1 angstrom (A) = 10 -10 m = 10~ 8 cm = 10 5 F It is noteworthy that the heaviest atoms are not markedly bigger than the lightest ones, although there are systematic variations, 27 Atoms Fig. 1-2 Relative atotnic radii (iiiferred from atomic volumes) versus atomic mass number, A. with pronounced peaks at the alkali atoms, as one progresses through the periodic table of the elements (see Fig. 1-2). Atoms are so small that it is hard to develop any real ap- preciation of the enormous numbers of atoms present in even the tiniest objects. For example, the smallest object that can be seen with a good microscope has a diameter of perhaps a few tenths of a micron anda mass of the order of 10~ ' 7 to 10~ ' fi kg. This minuscule object nevertheless contains something like 1 billion atoms. Or (to take another example) a very good labora- tory vacuum may contain residual gas at a pressure of a few times 10 -1 ' of atmospheric. One cubic centimeter of such a vacuum would likewise contain about 1 billion atoms. The atoms or molecules of a gas at normal atmospheric pressure are separated from one another, on the average, by about 10 times their diameter. This justifies (although only barely) the picture of a gas as a collection of particles that move independently of one another most of the time. MOLECULES; LIVING CELLS Our first introduction to molecules is likely to be in an elementary chemistry course, which very reasonably limits its attention to simple molecules made up of small numbers of atoms — H 2 0, C0 2) Na 2 S0 4 , C 6 H 6 , and the like, with molecular weights of the order of 10 or 100 and with diameters of a few angstroms. 28 A universe of particles Foot-and- mouth virus Edge of Bacterium coli Bushy stunt virus 10,000,000 Yellow fever virus' |Tobacco mosaic virus ( half-le ,000 Fig 1-3 Sizes of ■ Hemocyanin molecule 16,000,000 microscopic and sub- mo Hemoglobin molecule (63,000) microscopic objects, m A | bumin m0 |ecule (40,000) from bacteria down to _ Amino acjd chain _ 10 units (1|30 0) , . „ r, i t -a * Su 6 ar molecule (350) J. A. V. Butler, Inside _ ,,„„ % the Living Cell, ' Smgle am,n0 aC,d moleCule (130) George Allen & Vnwin, London, 1959.) 0.1 u.= 10,000 These then, do not represent much of an advance, either in size or in mass, on the individual atoms we have just been discussing. But through the development of biochemistry and biophysics we have come to know of molecules of remarkable size and com- plexity. We can feel justified in regarding them as particles on the strength of such features as a unique molecular weight for all molecules of a given type. The biggest objects that are de- scribable as single molecules have molecular weights of the order of 10 7 amu— hence masses of the order of 10~ 20 kg and lengths of the order of 10~ 7 m. Such objects are, however, far more important for their structure, and for their involvement in bio- logical processes, than for their rather precarious status as particles. The particle dynamics of a protein molecule is a pretty slim subject — limited perhaps to the behavior of the molecule in a centrifuge — whereas the elucidation of its structure is a study that requires (and merits) the most intensive efforts of brilliant chemists and crystallographers. It would be both presumptuous and inappropriate to attempt to discuss such matters here, but it is perhaps worth indicating the range of magnitudes of such particles with the help of Fig. 1-3. A convenient unit of length 29 Molecules; living cells for describing biological systems is the micron: 1 micron (m) = lO" 9 m = 10 4 A The largest object shown in Fig. 1-3 (a bacterium) is about 1m across and would be visible in a good microscope (for which the limit of resolution is about 0.2/n — rather less than one wavelength of light). Figure 1-3 includes some viruses, which have a peculiar status between living and nonliving — possessed of a rather definite size and mass, isolatable (perhaps as a crystalline sub- stance), yet able to multiply in a suitable environment. Figure 1-4 is an electron-microscope photograph of some virus particles. These are almost the smallest particles of matter of which we can form a clear image in the ordinary photographic sense. (You have perhaps seen "photographs" of atomic arrangements as observed with the device called a field ion microscope. These are not direct images of individual atoms, although the pattern Fig. 1-4 Sphericat particles of polio virus. [C. E. Schwerdt et al., Proc. Soc. Exptl. Biol. Med., 86, 310 (1954). Photograph courtesy of Robley C. Williams.] 30 A. universe of particles does reveal their spatial relationships.) 1 If we go one step further along this biological road, then of course we come to the living cell, which has the kind of sig- nificance for a biologist that the atom has for a physical scientist. Certainly it is appropriate to regard biological cells as particles, albeit of such a special kind that the study of most of them lies outside physics. They do, however, provide us with some con- venient reference points on our scale of physical magnitudes, and that is our only reason for mentioning them here — except, per- haps, for the matter of reminding ourselves that biological systems also belong within a framework defined by the fundamental atomic interactions. Although some single cells may be less than 1/* (certain bacteria) or more than 1 cm (e.g., the yolk of a hen's egg), the cells of most living organisms have diameters of the order of I0~ 5 m (1<V) on the average. Thus a human being, with a volume of about 0. 1 m 3 , contains about 10 ' 4 cells, each of which (on the average) contains about 10 l4 atoms. SAND AND DUST Vast areas of our earth arc covered with particles that have come from the breaking down of massive rock formations. These particles, predominantly of quartz (crystalline Si0 2 ), are chem- ically very inert and are just the kind of objects to which the word "particle" applies in ordinary speech — small but visible, and inanimate. The earth's surface, and the atmosphere above that surface, are loaded with such particles. The biggest ones (say, of the order of 1 mm across) rest more or less firmly on the ground. Others, orders of magnitude smaller, may be seen as motes of dust dancing in the sunlight, apparently showing no tendency to fail. The fate of a given particle depends on the combined effect of wind (or air resistance) and gravity. Ac- cumulations of windblown sand are found to be made up of particles from a maximum of about 1 mm diameter to a minimum of about 0.01 mm. Below that size the material tends to remain airborne, and the smallest dust particles are of the order of 'Improvemcnts in electron- and proton-microscope technique reach further and further down into the world of small particles. In 1969 the helical struc- ture of a protein molecule was photographed in an electron microscope. 31 Sand and dust Fig. 1-5 Size distribulion of particles found on or above the earlh's surface. (After R. A. Bagnold, The Physics of Blown Sand and Desert Dunes, Meihuen, London, 1941.) 10 -4 mm (=0.1/*) diameter. Figure 1-5 shows an approximate classification of particle sizes for dust, sand, and other small particles. OTHER TERRESTRIAL OBJECTS In a range from the smallest sand grains to the largest man-made objects we are in the realm of our most immediate experience: things that are large enough to be apprehended by the unaided sight or touch, yet not so big that we cannot achieve a rather direct awareness of them. In ordinary human terms, 100 is al- ready a large number, and a factor of 1000 up or down in linear scale from human dimensions brings us close to the limits of anything that can properly be regarded as a f uli, first-hand con- tact with the physical world. Outside that domain we depend chiefly upon indirect evidence, imagination and analogy. Since the densities of most solid materials (as we find them in the earth's crust) are of the order of magnitude of a few thousand kg/m 3 , the range of diameters from a grain of sand (1 mm) to a cliff or a dam (1 km) implies a range of masses from about 1 mg to 10' 2 kg (i.e., 10 9 tons). Actually, if mass itself (or the weight associated with it under the gravitational con- ditions at the earth's surface) were to be our criterion, then the range we have just stated goes far beyond the span of human perceptions. In terms of weights, it would be fair to say that our direct experience gives us some feeling for objects as light as about 10 mg or as heavy as 1000 kg— i.e., up to a factor of 10 ±4 with respect to a central value of the order of 0. 1 kg (or a few ounces). 32 A universc of particles PLANETS AND SATELLITES For an earthdweller going about his ordinary daily affairs, it is almost impossible— as well as unrealistic— to regard the earth as a particle. We cannot help but be aware, primarily, of the vast- ness of the earth and of the fact that man's greatest edifices are, at least in terms of physical size, totally insignificant modifica- tions of its surface. Yet if we can imagine ourselves backing off into space for 100 million miles or so, we can arrive at a very different point of view. The earth loses its special status and is placed on the same footing as the other planets. And all of them can, on this scale, be regarded as particles moving in their orbits about a much more massive particle, the sun. In Kepler's mathe- matical description of the orbits of the planets, and subsequently in Newton's dynamical theory of these motions, the planets could legitimately be regarded as point particles— their extent and internal structure were in no way relevant. The reason, of course, was the simple one that on the scale of the solar system the planets are little more than mere points, as is obvious whenever we look up into the night sky. The earth's diameter is only about 10 -4 of the distance between earth and sun — and the sun's own diameter is only 10 -2 of the same distance. No wonder, then, that a good first approximation to the dynamics of the solar system can be obtained by taking such ratios to be zero. But we should not rest content with such an approximation; indeed, we cannot. When Newton turned his attention from the planetary orbits to the tides, the physical extent of the earth became a key feature, because it was only this that made possible the existence of tide-producing forces, through a significant change of the moon's gravitational effect over a distance equal to the earth's diameter. In any case, once we free ourselves of the restriction of dealing with terrestrial objects and terrestrial phenomena, the particles that comprise the solar system are among the first to claim our attention. The planets can be described as roughly spherical particles, somewhat flattened as a result of their own rotation. Their range of diameters is considerable — a factor of nearly 30 between Mercury and Jupiter. Since the mean densities do not differ really drastically (the densest is our earth, at 5.5 times the density of water), the masses cover a very great range indeed. Again the extremes are Mercury and Jupiter, and there is a factor of about 6000 between their masses. Figure 1-6 depicts these principal 33 Planets and satellites Fig. 1-6 Relative sizes of llre planets and the sun. planets in correct relative scale. The size of the sun, which has about the same density as Jupiter but 10 times the diameter and about 1000 times the mass, is indicated for comparison. The same data are presented in Table 1-2. TABLE 1-2: DATA ON THE PLANETS Mean radius, km R/Re M/M E Mean density, kg/m 3 Mercury 2.42 X 10 3 0.38 0.054 5400 Venus 6.10 0.96 0.815 5100 Earth 6.37 1.00 1.000 5520 (Moon) (1.74) (0.27) (0.012) (3360) Mars 3.37 0.53 0.108 3970 Jupiter 69.9 10.98 317.8 1330 Saturn 58.5 9.20 95.2 680 Uranus 23.3 3.66 14.5 1600 Neptune 22.1 3.48 17.2 2250 Pluto 3.0 0.47 0.8? ? (Sun) (6.96 X 10 6 ) (1093) (3.33 X 10 5 ) (1400) The nine major planets represent almost all the mass of matter around the sun (and Jupiter alone accounts for almost two thirds of it), but the number of other captive objects is enormous. There are, first, the natural satellites. If we ignore man's contributions, there are about 30 known satellites of the planets, most of them extremely tiny in comparison with the planets to which they are tied. (Our own moon is relatively the largest, but even it has only a little over 1% of the mass of the earth.) These satellites have a very special dynamical in- terest, because from their motions it has been possible to infer the masses of the planets themselves. 34 A universe of particles STARS The planetary satellites are, however, vastly outnumbered by the minor planets, also called planetoids or asteroids. ' Tens of thousands of them are orbiting the sun in the region between the orbits of Mars and Jupiter. In size they range from 500 miles in diameter down to 1 mile or less, and they most probably come from the breakup of a larger body. If the astronomer Walter Baade's description of them — "the vermin of the skies" — is at all representative, they are not greatly beloved by professional skywatchers. A star is a magnificently complex structure, almost inconceivably gigantic by human standards and with a fascinating interior dynamics that involves nuclear reactions, gravitation, and elec- tromagnetism. Yet when we gaze out into space we see nothing of this, unless we look with the eye — and the instruments — of an astrophysicist. Instead, we see the stars as luminous points, which (in contrast to the planets) continue to appear as point sources when examined through even the most powerful tele- scopes. In relation to their diameters, in fact, stars are much farther away from each other than are the planets of our solar system. A convenient unit for specifying astronomical distances is the Hght-year, the distance light travels in a vacuum in 1 year: 1 light-year - 9.46 X 10 15 m The nearest star to the sun is about 4 light-years away, or about 25,000,000,000,000 miles! A number of other stars that are near neighbors of the sun have mutual separations of the order of 10 light-years or 10 17 m, which makes the ratio of diameter to separation about 10 -7 or 10 -8 . A cluster or galaxy of stars is thus an excellent example of a system of massive point particles, despite the fantastically large size of the particles. The best vacua attainable in the laboratory scarcely approach a corresponding emptiness as given by the ratio of interparticle spacing to particle diameter. One must look to the extreme vacua of interstellar space, where there are regions containing less than one hydrogen atom per cubic centimeter, to find a still emptier kind of system. •The word "asteroid" derives from the star-like appearance of these tiny objects as seen through a telescope. 35 Stars GALAXIES Most of the objects that are recognizable as stars are within the range of 10* 2 solar masses. In this sense our sun can be regarded as an average or typical star. In terms of size or lumi- nosity (total radiation) the range of variation is very much greater, but a worthwhile account of these features would really call for some discussion of the evolution and interior mechanism of stars; and this is certainly not the place to attempt it. 1 We shall content ourselves, therefore, with remarking that the stars, regarded simply as aggregations of matter, with masses between about 10 28 and 10 32 kg (thus containing something of the order of 10 54 to 10 58 atoms) can still be regarded as particles when we discuss their motions through space, because of the immense distances that separate them from one another. In 1900 the words "galaxy" and "universe" were regarded as being synonymous. Our universe appeared to consist primarily of a huge number of stars— many billions of them— scattered through space (see Fig. 1-1). Here and there, however, could be seen cloudy objects — nebulae — near enough or big enough to have an observable extent and even structure, as contrasted with the pointlike appearance of the stars. By 1900 many thousands of nebulae were known and catalogued. But what were they? To quote the astronomer Allan Sandage: "No one knew before 1900. Very few people knew in 1920. AH astronomers knew after 1924." 2 For it was in 1924 that the great astronomer Edwin Hubble produced the conclusive proof that the nebulae were, in the picturesque phrase, "island universes" far outside the region of space occupied by the Milky Way, and that our own Galaxy (distinguished with a capital G) was only one of in- numerable systems of the same general kind. The first suggestion for such a picture of the universe was in fact put forward by the philosopher Immanuel Kant as long ago as 1755, but of course at that time it was no more than a pure hypothesis. ; For extensive discussion, see (for example) F. Hoyle, Fronliers of Aslronomy, Harper & Row, New York, 1955. 2 The story of this development is fascinatingly told by Sandage in his iniro- duction to a beautifui book entitled The Hubble Alias of Galaxies, published in 1961 by the Carnegie Institution of Washington, D. C. See also Hubble's own classic work, The Realm of the Nebulae, Dover Publications, New York, 1958. 36 A universe of particles Fig. 1-7 Cluster of galaxies in the con- stellation Corona Borealis. Dislance about 600 million light-years. (Photo- graphfrom the Hale Observatories.) To quote Sandage again: "Galaxies are the largest single aggregates of stars in the universe. They are to astronomy what atoms are to physics." And so far as we can tell at present, they represent the largest particles in the observable scheme of things. A single galaxy may contain anywhere from about 10 6 to 10 ' ' stars. Our own Galaxy appears to be one of the larger ones, with a diameter of about 10 21 m (10 5 light-years). As we have already seen, the stars within an individual galaxy are very widely spaced indeed, so that the average density of matter in a galaxy is very, very low — only about 10~ 20 kg/m 3 . But even so, galaxies represent notable concentrations of matter. On the average the spacing between galaxies is about 100 times their diameter, although there is a tendency for them to exist in clusters with separations perhaps 10 times less than this. Thus even the galaxies may, on an appropriately large scale, be regarded as particles, and the interactions between them may be approximated by treating them as points (see Fig. 1-7). Astronomical surveys indicate that space contains a roughly uniform concentration of galaxies. If we assume that this con- tinues to be true up to the theoretical limits of observation, we can make some estimate of the content of the universe as a whole. For it would appear that the universe can be represented by a sphere with a radius of about 10 10 light-years (10 26 m). The 37 Galaxies PROBLEMS galaxies in general appear to be receding from us with speeds proportional to their distances, and at a distance of 10 l0 light- years the recessional speed would reach the speed of light. At this speed, because of the "galactic red shift" (a form of the wavelength and frequency shift for radiation from a moving source — called in general the Doppler effect), the transmission of energy back to us would fail to zero, thus setting a natural limit to the extent of the knowable universe. If we take the average density of matter throughout space to be about 10 -a(i kg/m 3 (10 -6 of the density within an individ- ual galaxy) and assume the total volume of the universe to be of the order of (10 26 ) 3 , i.e., 10 78 m 3 , we arrive at a total mass of about 10 52 kg (equivalent to about 10 79 hydrogen atoms). This would then correspond to a total of about 10 il galaxies, each containing about 10" stars. Such numbers are of course so stupendous that they defy any attempt to form a mental image of the universe as a whole. But it scems that we can be assured of one thing, at least, which is that the basis of our description of the physical world — what we described at the outset as "the par- ticulate view"— flnds some justiflcation over the entire range of our experience, from the nucleus to the cosmos. And the fact that this approach to the description of nature makes sense, while embracing a span of about 10 40 in distance and 10 80 in mass, is not merely aesthetically pleasing; it also suggests that something, at least, of the physical description we give. to the behavior of atoms will be found applicable to the behavior of galaxies, too. If we want to find a unifying theme for the study of nature, especially for the applications of classical mechanics, this par- ticulate view is a strong candidate. 1-1 Make a tabulation of the orders of magnitude of diameter, volume, mass, and density for a wide selection of objects that you regard as being of physical importance — e.g., nucleus, cell, and star. For the diameters and masses, rcprcsent the data by labclcd points on a straight line marked off logarithmically in successive powers of 10. This should give you a useful overview of the scale of the universe. 1-2 Estimate the number of atoms in: (a) The smallest speck of matter you can see with the naked eye. (b) The earth. 38 A universe of partieles 1-3 Calculate the approximate mass, in tons, of a teaspoonful of nuclear matter — closely packed nuclei or neutrons. 1-4 Sir James Jeans once suggested that each time any one of us draws a breath, there is a good chance that this lungful of air contains at least onc molecule from the last breath of Julius Caesar. Make your own calculation to test this hypothesis. 1-5 A vacuum pump evacuates a bottle to a pressure of 10~ 6 mm of mercury (about 10 -9 of atmospheric pressure). Estimate the magni- tudes of the following quantities: (a) The number of molecules per cubic centimeter. (b) The average distance between molecules. 1-6 In a classic experiment, E. Rutherford and T. Royds [Phil. Mag., 17, 281 (1909)] showed that the alpha particles emitted in radioactive decay are the nuclei of normal helium atoms. They did this by collecting the gas resulting from the decay and measuring its spectrum. They started with a source of the radioactive gas radon (itself a decay product of radium). The half-life of radon is 3.8 days; i.e., out of any given number of atoms present at any instant, half are left, and half have decayed, 3.8 days later. When the experiment started, the rate of alpha- particle emission by the radon was about 5 X 10 9 per second. Six days later, enough helium gas had been collected to display a complete helium spectrum when an electric discharge was passed through the tube. What volume of helium gas, as measured at STP, was collected in this experiment? [The number of surviving radioactive atoms as a function of time is given by the exponential decay law, N(t) = Noe-*'. The number of disintegrations per unit time at any instant is \dN/dt\ = \Noe- u . Given the value of the half-life, you can deduce the value of the constant, X.] 1-7 The general expansion of the universe as described by Hubble's law [o{r) = ar] implies that the average amount of mass per unit volume in the universe should be decreasing by about 1 part in 10 17 per second (see Hors d'oeuvres no. 29, p. 16). According to one theory (no longer so strongly held) this Ioss is being made up by the continuous creation of matter in space. (a) Calculate the approximate number of hydrogen atoms per cubic meter per year that would, if produced throughout the volume of the universe, bring in the necessary amount of new mass. (b) It has been hypothesized (by J. G. King) that the creation of new matter, if it occurs, might well take place, not uniformly throughout space, but at a rate proportional to the amount of old matter present within a given volume. On this hypothesis, calculate the number of hydrogen atoms equivalent to the new mass created per day in a vessel containing 5 kg of mercury. (J. G. King concluded 39 Problcms on this basis that a test of the hypothesis was feasible.) Convert your result to an equivalent volume of hydrogen gas as measured at a pres- sure of 1 Torr (1 mm of mercury). 1-8 Theodore Rosebury, in his book Life on Man (Viking, New York, 1969), remarks that the total bulk of all the microorganisms that live on the surface of a human being (excluding a far larger quantity on the inner surfaces of the intestinal system) could easily be accommodated in the bottom of a thimble. If the mean radius of these microorganisms is taken to be 5 y. (1 n = 10 -6 m), what is the order of magnitude of the population of such organisms that each of us carries around all the time ? Compare the result with the total human population of the globe. 1-9 It seems probable that the planets were formed by condensation from a nebula, surrounding the sun, whose outer diameter corresponded roughly to the orbit of Pluto. (a) If this nebula were assumed to be in the form of a disk with a thickness equal to about one tenth of its radius, what would have been its mean density in kg/m 3 ? (Do not include the mass of the sun in the calculation.) (b) A better picture appears to be of a gas cloud whose thickness increases roughly in proportion to the radius. Suppose that Jupiter were formed from a ring-shaped portion of the cloud, extending radially for half the distance from the asteroid belt to the orbit of Saturn and with a thickness equal to this radial extension. What would have been the mean density of this portion of the nebula ? (In terms of the radius of the earth's orbit, the orbit of Pluto has a radius of about 40 units, the asteroid belt is at about 3 units, and the orbit of Saturn has a radius of about 9.5 units.) 1-10 The core of a large globular cluster of stars may typically contain about 30,000 stars within a radius of about 5 light-years. (a) Estimate the ratio of the mean separation of stars to the stellar diameter in such a core. (b) At approximately what degree of vacuum would the same ratio of separation to particle diameter be obtained for the molecules of a gas? 40 A universe of particles / do not define time, space, place and motion, since they are well known to all. newton, Principia (1686) Space, time, and motion WHAT IS MOTION? you are undoubtedly familiar with motion in all kinds of mani- festations, but what would you say if you were asked to define it? The chances are that you would find yourself formulating a statement in which the phrase "a change of position with time," or something equivalent to that, expressed the central thought. For it seems that our ability to give any precise account of motion depends in an essential way on the use of the separate concepts of space and time. We say that an object is moving if it occupies different positions at different instants, and any stroboscopic photograph, such as that shown in Fig. 2-1, gives vivid expression to this mental picture. Ali of us grow up to be good Newtonians in the sense that our intuitive ideas about space and time are closely in harmony with those of Newton himself. The following paragraphs are a deliberate attempt to express these ideas in simple terms. The description may appear natural and plausible, but it embodies many notions which, on closer scrutiny, will turn out to be naive, and difficult or impossible to defend. So the account below (set apart with square brackets to emphasize its provisional status) should not be accepted at its face value but should be read with a healthy touch of skepticism. [Space, in Newton's view, is absolute, in the sense that it exists permanently and independently of whether there is any matter in the space or moving through it. To quote Newton's own words in the Principia: "Absolute space, in its own nature, 43 Fig. 2-1 Slrobo- scopic photograph of a motion. {Front PSSC Physics, D. C. Healh, Lexington, Massachusetis, 1965.) without relation to anything external, remains always similar and immovable." [Space is thus a sort of stationary three-dimensional matrix into which one can place objects or through which objects can move without producing any interaction between the object and the space. Each object i n the universe exists at a particular point in space and time. An object in motion undergoes a continuous change of its position with time. And although it would not be practicable, one can imagine the charting of positions with the help of a vast network of meter sticks, laid out end to end in a three-dimensional, cubical array throughout space. One can conceive of extending such measurements to any point in the universe. In other words, the space is there, and we simply have a practical task of attaching markers to it. Moreover, our physical measurements agree with the theorems of Euclidean geometry, and space is thus assumed to be Euclidean. [Time, in Newton's view, is also absolute and flows on with- 44 Space, time, and motion out regard to any physical object or event. Again quoting from the Principia: "AbsoLute, true, and mathematical time, of itself, and from its own nature, flows equably without relation to any- thing external, and by another name is called duration." The language is elegant but delightfully uninformative. As Newton said in the remark quoted at the beginning of this chapter, he did not attempt to define either space or time. [One can neither speed up time nor slow down its rate, and this flow of time exists uniformly throughout the universe. If we imagine the instant "'now" as it occurs simultaneously on every planet and star in the universe, and an hour later mark the end of this 60-minute interval, we assume that such a time interval has been identical for every object in the universe, as could (in principle) be verified by observations of physical, chemical, or biological processes at various locations. As an aid to measuring time intervals, it would be possible, in principle, to place identical clocks at each intersection of a meter-stick framework and to synchronize these clocks so that they indicate the same time at a common, simultaneous instant. Being identical clocks, they would thereafter correctly mark off the flow of absolute time and remain synchronized with each other. [Space and time, although completely independent of each other, are in a sense interrelated insofar as we find it impossible to conceive of objects existing in space for no time at all, or existing for a finite time interval but "nowhere" in space. Both space and time are assumed to be infinitely divisible — to have no ultimate structure.] The preceding five paragraphs describe in everyday language some commonsense notions about the nature of space and time. Embedded in these notions are many assumptions that we adopt, either knowingly or unconsciously, in developing our picture of the universe. It is fascinating therefore that, however intuitively correct they seem, many of these ideas have consequences that are inconsistent with experience. This first became apparent (as we mentioned in the Prologue) in connection with motions at very high speeds, approaching or equaling the speed of light, and with the phenomena of electromagnetism ; and it was Einstein, in his development of special relativity theory, who exposed some of the most important limitations of classical ideas, in- cluding Newton's own ideas about relativity, and then showed how they needed to be modified, especially with regard to the concept of time. 45 What is motion '? The crux of the matter is that it is one thing to have abstract concepts of absolute space and time, and it is another thing to have a way of describing the actual motion of an object in terms of measured changes of position during measured intervals of time. Newton himself understood this very well, at least as far as spatial measurements were concerned. Thus in the Principia we find him remarking: "But because the parts of space cannot be seen, or distinguished from one another by our senses, there- fore in their stead we use sensible [i.e., observable] measures of them . . . And so, instead of absolute places and motions, we use relative ones . . . For it may be that there is no body really at rest, to which the places and measures of others may be re- ferred." If there is any knowledge to be gained about absolute space, it can only be by inference from these relative measure- ments. Thus our attention turns to the only basis we have for describing motion — observation of what a given object does in relation to other objects. FRAMES OF REFERENCE If you should hear somebody say "That car is moving," you would be quite certain that what is being described is a change of position of the car with respect to the earth's surface and any buildings and the like that may be nearby. Anybody who an- nounced "There is relative motion between that car and the earth" would be rightly regarded as a tiresome pedant. But this does not alter the fact that it takes the pedantic statement to express the true content of the colloquial one. We accept the local surroundings — a collection of objects attached to the earth and therefore at rest relative to one another— as defining a frame ofreference with respect to which the changes of position of other objects can be observed and measured. It is clear that the choice of a particular frame of reference to which to refer the motion of an object is entirely a matter of taste and convenience, but it is often advantageous to use a reference frame in which the description of the motion is simplest. A ship, for example, is for many purposes a self-contained world within which the position or path of any person is most efficiently described in terms of three perpendicular axes based on the directions fore and aft, port and starboard, and up and down (according to deck number). The uniform motion of the ship itself 46 Space, time, and motion with respect to a frame of reference attached to the earth may be unnoticed or even ignored by the passengers, as long as they can rely on the navigation officers. Is the earth itself at rest? We would not say so. We have become accustomed to the fact that the earth, like the other planets, is continually changing its position with respect to a greater frame of reference represented by the stars. And since the stars constitute an almost completely unchanging array of Fig. 2-2 Apparent circular molions of the stars. (Carillon Tower of Wellesley College, Wellesley, Mass., from J. C. Duncan, Astronomy, 5th edilion, Harper &Row,J954,p. 19.) 'i» • • ' ' JK f' ] ' '■ - i T , ir f f 'f f %ti<$ ntjaKKm 47 Frames of reference reference points in the sky, we regard the totality of them as representing a fixed frame of reference, within which the earth both rotates and moves bodily. It remains true, however, that our primary data are only of relatioe positions and displacements; the belief that it makes more sense to assume that the earth is really rotating on its axis once every 24 hours, rather than that the system of stars is going around us, is one that cannot be justified by primary observations alone (see Fig. 2-2). With our present knowledge of the great masses and enormous numbers of the stars, it does, to be sure, simply seem more reasonable to attribute the motion to our puny earth. But it would be hard to elevate that subjective judgment into a physical law. Later we shall see that there are powerful theoretical reasons for preferring some reference frames to others. The "best" choice of reference frame becomes ultimately a question of dynamics — i.e., dependent on the actual laws of motion and force. But the choice of a particular reference frame is often made without regard to the dynamics, and for the present we shall just concern ourselves with the purely kinematic problems of analyzing positions and motions with respect to any given frame. COORDINATE SYSTEMS A frame of reference, as we have said, is defined by some array of physical objects that remain at rest relative to one another. Within any such frame, we make measurements of position and displacement by setting up a coordinate system of some kind. In doing this we have a free choice of origin and of the kind of coordinate system that is best suited to the purpose at hand. Since the space of our experience has three dimensions, we must in general specify three separate quantities in order to fix uniquely the position of a point. However, most of the problems that we shall consider will be of motion confined to a single plane, so let us first consider the specification of positions and displacements in two dimensions only. As you are doubtless well aware, the position of a point in a plane is most often designated with respect to two mutually perpendicular straight lines, which we call the x and y axes of a coordinate system, intersecting at an origin O. The position of the point P [Fig. 2-3(a)] relative to O is then described by a position vector r, as shown, characterized by a specific length and 48 Space, time, and motion Fig. 2-3 (a) Square grid; the basis of Carlesian coordinates in a plane. (b) Plane polar coordinate grid. a specific direction. Using our perpendicular axes, we can uniquely define r by the pair of rectangular (Cartesian) co- ordinates (x, y), which are projections of r onto the x and y axes, respectively. However, another important way of specify- ing r is in terms of polar coordinates (r, 0), as shown in Fig. 2-3(b). Here r is the distance of P from O and d is the angle that r makes with the positive x axis, as measured in a counterclock- wise (conventionally positive) direction. The two schemes of designating the position of P are related as follows: (In two dimcnsions) r 2 = x 2 + y 2 y tan = - X x = r cos d y = = r sin (2-1) Figure 2-4 shows examples of the use of these coordinate systems. We shall on various occasions be making use of unit vectors that represent displacements of unit length along the basic co- ordinate directions. In the rectangular (Cartesian) system we shall denote the unit vectors in the x and y directions by i and j, respectively. The position veetor r can be written as the sum of its two veetor components: (In two dimensions) r = xi + y\ (2-2) In the polar coordinate system, we shall use the symbol e r to denote a unit veetor in the direction of increasing r at constant 6 and the symbol e e to denote a unit veetor at right angles to r in the direction of increasing 9. (The use of the symbol e for this purpose comes from the German word for unit, which is 49 Coordinate systems /*>" '""'V (a) y>«iui.i«ns< * (b) fig. 2-4 (a) Example of Cariesian coordinales in use — a portion of mid- town Manhattan, New York City. (b) Example of plane polar coordinales in use— a radar scopeface with afew incipient thunderslorms (June 3, 1970). North is shown as O" azimuth. The heavy circles ofr = const are at 100 km, 200 km; the lighter circles are spaced by 25 km. (Photograph courtesy of Department of Meteorology, M.l.T.) "Einheit.") In this polar-coordinate system, the vector r is simply equal to re r , and one might wonder why the unit vector e« is introduced at all. As we shall see, however, it becomes very important as soon as we consider motions rather than static displacements, for motions will often have a component per- pendicular to r. Although the above coordinate schemes are the most familiar ones — and are the only ones we shall be using in this book for two-dimensional problems — it is worth noting that any mapping of the surface that uniquely fixes the position of a point is a possible system. Figure 2-5 shows two examples — one a non- orthogonal system based on straight axes, and the other an orthogonal system based on two sets of intersecting curves. Such systems are introduced to capitalize on the kinds of symmetry that particular physical systems may possess. 50 Spacc, time, and motion Fig. 2-5 (a) Oblique coordinate system, ofthe kind Ihal is freguently used in Minkowski diagrams (space- time diagrams) in special relativity. (b) Orthogonal curvilinear coordinate system, made o/ intersecting sets of confocal ellipses and hyperbolas. If it is necessary to specify all three of the spatial coordinates of a point, the most generally useful coordinate systems are the three-dimensional rectangular (Cartesian) coordinates (x, y, z), and spherical polar coordinates (r, 6, <p). These are both illus- trated in Fig. 2-6(a). The Cartesian system is almost always chosen to be right-handed, by which we mean that the positive z direction is chosen so that, looking upward along it, the process of rotating from the positive x direction toward the positive y direction corresponds to that of a right-handed screw. It then follows that the cyclic permutations of this operation are also right-handed — from -\-y to +z, looking along +x, and from +z to +x, looking along +y. You may note that the two- dimensional coordinate system, as shown in Fig. 2-3(a), would on this convention bc associated with a positive z axis sticking up toward you out of the plane of the paper. Introducing a unit vector k in the +z direction, analogous to the i and j of Eq. (2-2), we can write the vector r as the sum of its three Cartesian vector components: r = xi + yi + zk 51 Coordinate systems (2-3) Fig. 2-6 (a) Coordinates ofa poinl in three dimensions, showing both spherical polar and right-handed Carlesian coordinates. (6) Point located by angular coordinates (lalitude and longitude) on a sphere, and the unit vectors ofa local Carlesian coordinate system at the point in question. The description of the position or displacement in spherical polar coordinates makes use of one distance and two angles. (Notice that three dimensions requires three independent co- ordinates, whatever particular form they may take.) The distance is, as with plane polar coordinates, the distance r from the chosen origin. One of the angles [the one shown as in Fig. 2-6(a)] is simply the angle between the vector r and the positive z axis; it is known as the polar angle. The other angle represents the angle between the zx plane and the plane defined by the z axis and r. It can be found by drawing a perpendicular PN from the end point P of r onto the xy plane and measuring the angle be- tween the positive x axis and the projection ON. This angle (p) is called the azimuth. The geometry of the figure shows that the rectangular and spherical polar coordinates are related as follows: x = r sin cos tp y = r sin sin <p z = r cos (2-4) If we set = tt/2, we make z = and so get back to the two- 52 Space, time, and motion dimensional world of the xy plane. The first two equations of (2-4) then give us x = r cos ¥> y = r sin v? It is very unfortunate that a long-established tradition uses the symbol d, as we ourselves did earlier, to denote the angle between the vector r and the x axis in this special two-dimensional case. This need not become a cause for confusion, but one does need to be on the alert for the inconsistency of these conventions. We have all grown up with one important use of spherical polar coordinates, the mapping of the earth's surface. This is indicated in Fig. 2-6(b). The longitude of a given point is just the angle ip, and the latitude is an angle, X, equal to jr/2 — 6. (This entails calling north latitudes positive and south latitudes negative.) At any given point on the earth's surface a set of three mutually orthogonal unit vectors defines for us a local coordinate system; the unit vector e r points vertically upward, the vector e e points due south, parallel to the surface, and the third unit vector, e^,, points due east, also parallel to the surface. As with the plane polar coordinates, the vector r is given simply by re r . COMBINATION OF VECTOR DISPLACEMENTS Suppose we were at a point P\ on a flat horizontal plane [Fig. 2-7 (a)] and wished to go to another point P g. Imagine that we chose to make the trip by moving only east and north (represented by +x and +y in the figure). We know there are two particularly Fig. 2-7 (a) Suc- cessiue displacements on a plane; the final position is independen! of the order in which the displacements are made. {b) Addition of secerat displace- ment vectors in a plane. 53 Combination of vector displacements ■■■■ Fig. 2-8 Seatar multiples of a given veetor r, including negative multiples. X 2r IHnBBHBBHHBBHBB^HEHBHiBI straightforward ways of doing this: (1) travel a certain distance s z due cast and then a certain distance s u due north, or (2) travel s u due north, followed by s x due east. The order in which we take these two component displacements does not matter; we rcach the same point P 2 in either casc. Our representation of the veetor r in Eq. (2-2) as the sum of the individual veetors xi and yj is an example of such a combination. This simple and familiar property of linear displacements exemplifies an essential feature of all those quantities we call veetors and is not confined to combinations at right anglcs. Thus, for example, in Fig. 2-7(b) we illustrate how three veetor displacements, A, B, and C, placed head to tail, can be combined into a single veetor displacement S drawn from the original starting point to the final end point. This is what we mean by adding the veetors A, B, and C. The order in which veetors are added is of no consequence; thus successive displacements of an object can be combined according to the veetor addition law, without regard to the sequence in which the displacements are made. What we mean by a veetor quantity, in general, is that it is a direeted quantity obeying the same laws of combination as positional displacements. We shall often be concerned with forming a numerical multiple of a given veetor. A positive multiplier, n, means that we changc the length of the veetor by the faetor n without chang- ing its direetion. The negative of a veetor (multiplication by — 1) is defined to mean a veetor of equal magnitude but in the opposite direetion, so that added to the original veetor it gives zero. A negative multiplier, —n, then defines a veetor reversed in diree- tion and changed in length by the faetor n. These operations are illustrated in Fig. 2-8. Subtracting one veetor from another is accomplished by noting that subtraetion basically involves the addition of a nega- tive quantity. Thus if veetor B is to be subtraeted from veetor A, we form the veetor —B and add it to A: A - B = A I (-B) 54 Space, time, and motion Fig. 2-9 Addilion and subtraction of Iwo given vectors. Note that the magni- tude ofthe vector difference may be {as here) larger than the Fig. 2-10 Construc- tion of the relative position vector ofone point (i> 2 ) with re- spect to another point OPi). In Fig. 2-9 we show both the sum and the difference of two given veetors. We have deliberately chosen the direetions of A and B to be such that the vector A — B is longer than the vector A + B; this will help to emphasize the faet that vector combination is something rather different from simple arithmetical combination. The evaluation of the vector distance from a point P^ to a point P 2 , when originally the positions of these points are given separately with respect to an origin O [see Fig. (2-10)], is a direct application of vector subtraction. The position of P 2 relative to P\ is given by the vector r 12 such that ri2 = r2 — ri (The subseript "12" is to be read as "one-two" and is a common notation in the deseription of two-particle systems.) Similarly, the position of P! relative to P 2 is given by the vector r 2 i = r, — r 2 . Clearlyr 2 i = — r 12 . 55 Combination of vector displacements Fig. 2-11 (a) Com- ponents of a given vector in two different rectangular coordi- nate systems related by an angular dis- placemenl 9 in the xy plane. (b) Com- ponents of a given vector in a rectangu- lar coordinate system and in an obligue co- ordinate system. THE RESOLUTION OF VECTORS 1 In discussing the description of a given vector in terms of its components, we have indicated that this is a process that can be operated in either direction. There is the analysis (resolution) of the vector into its components in a given coordinate system, or there is the synthesis (addition) of the vector components to reconstitute the original vector. There is, however, an important difference between thcse two operations. The vector sum of the components is unique— it is the particular vector that we are considering — but the process of resolving the vector into com- ponents can be done in an infinity of ways, depending on the choice of coordinate system. If we are using a coordinate system based on orthogonal axes (whether Cartcsian or polar or anything else), the com- ponents of the vector are easily found by multiplying the length of the vector by the cosine of the angle that the vector makes with cach of the coordinate axes in turn. Thus, for example, if we had a vector A confined to the xy plane [see Fig. 2-1 l(a)] its components in the xy coordinate system are given by A x = Acosa A„ = /4cos<3 We know that /3 = (tt/2) - «> but by introducing the two separate angles we have a formalism that lends itself to being 'U is hoped that this seetion may be helpful in a general way, but the only fcaturc that will be specifically needed later is the scalar produet of two arbi- trary veetors. 56 Space, (ime, and motion extended to the case of three dimensions, making use of the three separate angles a, 0, and T that the vector makes with the three axes. In this more general case there is no simple connection between the angles themselves, but we have the relationship cos 2 a + cos 2 /3 + cos 2 7 = 1 Our two-dimensional case corresponds to putting 7 = w/2. The total vector A in Fig. 2- 11 (a) can of course be written A = AJ + A v \ = {A cos d)\ + (A cos ;8)j A very convenient way of expressing such results is made possible by introducing what is called the scalar product of two vectors. This is defined in general in the following way: If the angle be- tween any two vectors, A and B, is 0, then the scalar product, S, is equal to the product of the lengths of the two vectors and the cosine of the angle d. This product is also called the dot product because it is conventionally written as A • B. Thus we have scalar product (S) = A ■ B = AB cos d If for the vector B we now choose one or other of the unit vectors of an orthogonal coordinate system, the scalar product of A with the unit vector is just the component of A along the direction characterized by the unit vector: A x = A ■ i A y = A • j Thus the vector A can be written as follows: A = (A-i)i + (A-j)j This result can, in fact, be developed directly from the basic statement that A can be written as a vector sum of components along x-and y: A = A x \ + Ay\ Forming the scalar product of both sides of this equation with the unit vector i, we have A • i = A x (i ■ i) + A„(i ■ i) Now (i • i) = 1 and (j • i) = 0, because these vectors are all of unit length and the values of are and t/2, respectively. Thus 57 The resolution of vectors we have a more or less automatic procedure for selecting out and evaluating each component in turn. If one were to take this no further, the above development would seem perhaps pointlessly complicated. Its value becomes more apparent if one is interested in relating the components of a given vector in different coordinate systems. Consider, for example, the second set of axes (x\ y') shown in Fig. 2-1 1 (a) ; they are obtained by a positive (counterclockwise) rotation from the original (x,y) system. The vector A then has two equally valid representations: A - AJL + A„j = AJV + A u '¥ If we want to find A» in terms of A x and A v , we just form the scalar product with i' throughout. This gives us A,' = AS • i') + A„(i ■ i') Looking at Fig. 2-1 l(a), we see the following relationships: i • i' = cos 6 j • i' = cos f | - 6 j = sin Hence A z ' = A z cos + A„ sin Similarly, A y ' = AA-f) + AJl'f) = /*xCOs(;; + B ) + /*„COS0 Therefore, A y ' = -A x sin d + A y cos 6 This procedure avoids the need for tiresome and sometimes awkward considerations of geometrical projections of the vector A onto various axes of coordinates. The same approach can be useful if a vector is to be resolved into nonorthogonal components. Consider, for example, the situation shown in Fig. 2-1 1 (b). The axes are the lines Osi and Os 2 , and the components of the vector A in this system are OE and OF. If we denote unit veetors along the coordinate direetions by ei and e 2 , we havc 58 Space, timc, and motion A = siei + S2*2 = A z i + A„\ If we form the scalar product throughout with, let us say, ei, we havc s\ + J2(e2 • ei) = A x (i ■ ei) + /i„(j • ci) Given a knowledge of the angles between the various axes, this is a linear equation involving the two unknowns Si and s 2 . A second equation can be obtained by forming the scalar product throughout with e 2 instead of ei, and it then becomes possible to solve for Si and s 2 separately. It is important, in this case, to recognize that the base vectors e! and e 2 are no longer or- thogonal, so that the scalar product (e, • e 2 ) does not vanish. The use of oblique coordinate systems of this kind is, however, rather special, and as a rule the resolution of a vector along the three independent directions of an orthogonal coordinate system is the reasonable and useful thing to do. VECTOR ADDITION AND THE PROPERTIES OF SPACE 1 You may be tempted to think that the basic law of vector addi- tion, and the fact that the final result is independent of the order in which the combining vectors are taken, is more or less obvious. Let us therefore point out that it dcpends crucially on having our space obey the rules of Euclidean geometry. If we are dealing with displacements confined to a two-dimensional world as represented by a surface, it is essential that this surface be flat. This is not just a pedantic consideration, because one of our most important two-dimensional reference frames — the earth's surface — is curved. As long as displacements are small compared to the radius of the curvature, our surface is for practical pur- poses flat, our observations conform to Euclidean plane geometry, and all is well. But if the displacements along the surface are sufliciently great, this idcalization cannot be used. For cxample, a displacement 1000 miles eastward from a point on the equator, followed by a displacement 1000 miles northward, does nol bring one to the same placc as two equivalent displacements (i.e., again on grcat circles intersecting at right angles) taken in ■This scction can be omitted without loss of continuity. 59 Vector addition and the properties of space Fig. 2-12 Successiue displacements on a sphere are not com- mutative if the sizes of the displacements are not small com- pared to the radius of the sphere. the opposite order; it misses by about 40 miles! (See Fig. 2-12.) This means, in effect, that the correctness of vector addition is a matter for experiment and that tests for departures from it can be used to make deductions about the geometrical properties of the space in which we operate. For example, we could take the results of sensitive measurements on the difference between the two possible ways of making two successive displacements on a sphere and use the data to deduce the radius of the sphere. When we acknowledge that the space of our ordinary ex- perience is really three-dimensional, then, of course, we look at the foregoing analysis from a different point of view. We recog- nize that displacements on the surface of a sphere can actually be scen as displacements in a three-dimensional world that obeys Euclidean geometry rather than in a merely two-dimensional world that appears, within itself, to be non-Euclidean. But at this point a very intercsting speculation suggests itself: Can we say that our space of three dimensions is rigorously Euclidean? Is it possible that the result of adding displacements along the three basic coordinate dircctions is dependent to some minute extent on the order of addition? If this were discovered to be the case, then we might proceed by analogy and introduce a fourth spatial dimension, associated with some characteristic radius of curvature, such that our non-Euclidean space of three dimensions could be described as Euclidean in a "hyperspace" of four dimensions. 60 Space, tirae, and motion TIME Some distinguishcd scientists, beginning with the great Kari Friedrich Gauss ("the Prince of mathematicians") ' have sought by direct observation to test the validity of Euclidean geometry, by measuring whether the angles of a closed triangle add up to exactly 180°. (For example, in the non-Euclidean space repre- sented by the surface of a sphere, the angles of a triangle add up to more than 180°.) No departure from a Euclidean character for three-dimensional space has been detected through such observa- tions. The concept that space may, however, be "curved," and that this curvature might be revealed if one could only carry out observations over sufficiently great distances, occupies an im- portant place in thcoretical cosmology. You may have read about the curvature of space in another connection— Einstein's theory of gravitation— which describes local gravitational effects in terms of a modification of geometry in the space surrounding a massive object such as the sun. We shall not pursue this topic here, although we shall touch on it very briefly at the end of our account of gravitation in general (Chapter 8). In the preceding sections we have developed the basic analysis of spatial displacements. To describe motion we must link such displacements to the time intervals during which they occur. Before considering this as a quantitative problem, let us very briefly supplement the remarks that we made at the beginning of the chapter concerning the actual nature of time. This is, of course, a hugc subject that has engaged the thoughts and specula- tions of men— philosophers, scientists, and humanists alike— throughout history, and continues to do so. We shall not presume to do more than to cxamine one or two aspects of the problem from the standpoint of physical seienec. The sense of the passage of time is deeply embedded in every one of us. We know, in some elemental sense, what time is. 'Kari Friedrich Gauss (1777-1855) was one of the outstanding mathema- ticians of all time. In the originality and range of his work he has never been surpassed, and probably never equaled. He delved deeply into astronomy and geodesy and was perhaps the first to recognize the possibility of a non- Euclidean geometry for space. See E. T. Bell's essay about him in The World of Mathemalics (J. R. Newman, ed.), Simon and Schuster, New York, 1956. 61 Time But can we say what it is? The distinguished Dutch physicist H. A. Kramers once remarked: "My own pet notion is that in the world of human thought gencrally, and in physical science particularly, the most important and most fruitful concepts are those to which it is impossible to attach a well-defined meaning." ' To nothing, perhaps, does this apply more cogently than to time. Nevertheless, if one tries to analyze the problem, one can perhaps begin to see that it is not entirely elusive. Even though a defini- tion of time may be hard to come by, one can recognize that our concept of the passage of time is tied very directly to the fact that things change. In particular we are aware of certain recurrent events or situations — the beats of our pulse, the daily passage of the sun, the seasons, and so on. We almost subconsciously treat these as though they are markers on some continuous line that already exists— rather like milestones along a road. But all we have in terms of direct knowledge is the set of markers; the rest is an intellectual construction. Thus, although it may be valuable to have an abstract concept of time continuously flowing, our first-hand experience is only the observed behavior of a device called "a clock." In order to assign a quantitative measure to the duration of some process or the interval between two events, we simply associatc the beginning and end points with readings on a clock. It is not essential that the clock make use of a recurrent phenomenon — one need only consider ancient devices such as water clocks and graduated candles, or modern parallels such as continuously weakening radioactive sources — but we do ask that it provide us with a means of marking off successive intervals in some recognizable way, and most such devices do, in fact, make use of repetitive phenomena of some kind. How do we know that the successive time intervals defined by our chosen clock are truly cqual? The fact is that we don't; it is ultimately a matter of faith. No clock is perfect, but we have learned to recognize that some clocks are better than others — better in the sense that the segments into which they divide our experience are more nearly equal. A doctor observes his wristwatch and tells us that our pulse is irregular; the wrist- watch, however, is found to be itself irregular when compared 'Physical Sciences and Human Values (a symposium), Princeton University Press, Princeton, N. J., 1947. 62 Space, time, and motion more critically against a crystal-controlled oscillator; the oscil- lator wandcrs noticeably when checked against a clock based on atomic vibrations. Whether or not the uniform flow of time, as an ultimate abstraction, has any physical meaning, the remarkable fact is that we approach the measurement of time as if this steady flow existed. We evaluate the behavior of any given clock by observing its consistency and reproducibility, so that in using it we can quote its measure of a time or a time interval with a specified range of possible error. And then, when we proceed from individual measurements to general equations involving time, we introduce the symbol t and treat r as a continuous variable in the mathematical sense. UNITS AND STANDARDS OF LENGTH AND TIME Most of our discussion of motion will be in terms of unspecified positions and times, represented symbolically by r, /, and so on. It should never be forgotten, however, that the description of actual motions involves the numerical measures of such quan- tities and the use of universally accepted units and standards. Our choice of acceptable standards of both distance and time is the result of a continuing search for the highest degree of con- sistency and reproducibility in such measurements. The evolution and present state of this process is briefly summarized below. Length The current Standard of length — the meter — was introduced, along with the rcst of the metric system, in the drive for scientific and cultural order that developed in France in the latter half of the 18th century. The meter was originally intended to repre- sent 1 ten-millionth (10 -7 ) of the distance from pole to equator of the earth along a meridian of longitude. But it proved im- possible to construct any sufficiently precise Standard on the basis of this definition, and the meter was then defined as the length of a particular metal bar kept in Sevres, near Paris, and finally as a multiple of the wavelength of a charaeteristie spectral line — most recently (since 1960) as being a distance equal to 1,650,763.73 wavelengths of orange-red light from the isotopc krypton 86. Although the use of the extremely small units of length 63 Units and standards of length and time represented by light waves means that the meter can be defined with immense precision — to 1 part in 10 8 or better — there is the disadvantage that an object as long as a meter cannot be directly measured, in a one-step process, in terms of light waves from ordinary sources. The reason is that the measurements depend on observing optical interference effects that begin to wash out if the distance in question becomes more than about 1 ft. The development of lasers has completely transformed this situation, and interference effects have been observed up to path lengths of over 100 m. It thus seems quite probable that the meter will at some future date be defined in terms of an optical wavelength obtainable from a laser source, perhaps one of the characteristic spectral lines of neon in a helium-neon laser. Time The process of defining a Standard of time involves a feature that sets it significantly apart from the establishment of a Standard of length. This is that, as Allen Astin has remarked: "We cannot choose a particular sample of time and keep it on hand for reference." 1 We depend upon identifying some recurring phe- nomenon and assumirrg that it always supplies us with time intervals of the same length. The Standard of time — the second — was originally based on the assumed constancy of the earth's rotation. It was first defined as being equal to 1/86,400 of a mean solar day — i.e., the average, over 1 year, ol the time from noon to noon or midnight to mid- night at a given placc on the earth's surfacc. This is an awkward definition, because the length of the day, as measured from noon to noon, is not a constant; it varies because the earth's speed and its distance from the sun are continuously changing during one complete orbit. A logically more satisfaetory definition of the second can be based on the sidereal day— the time for any given star to return to the same position overhead. If the earth's rotation were truly uniform, the length of every sidereal day would be the same. In faet, it has gradually come to be rccognized, thanks to 'Allen V. Astin, "Standards of Measurement," Sci. Am., 218 (6), 50 (1968). Some people would, however, argue that even a solid, tangible bar as a length Standard is equally vulnerable on philosophical and logical grounds. 64 Space, time, and motion the extraordinary precision of astronomical raeasurements ex- tended over thousands of years, that neither the length of the solar year nor the earth's rate of rotation on its axis is exactly constant, the latter in particular being subject to minute but abrupt variations. The year has been found to be lengthening at the rate of about £ sec per century, so that in 1956 the second was redefined as being equal to 1/31,556,925.9747 of the tropical year 1900. (A tropical year is defined as the interval of time between two successive passages of the sun through the vernal equinox. We shall not attempt here to deseribe just how one "latehes on" to a second of the year 1900 for calibration pur- poses.) It should be recognized that the variations being dis- cussed here are fantastically small, as is implied by the ability to define the year in terms of the second to 12 significant figures. Finally, in 1967, the use of atomic vibrations to specify a time Standard was adopted by international agreement ; it defines the second as corresponding to 9,192,631,770 eyeles of vibration in an atomic clock controlled by one of the charaeteristie fre- quencies associated with atoms of the isotope cesium 133. Quite apart from the practical challenge of defining units and standards of time with the maximum attainable precision, there are some questions of fundamental interest involved. Does time as defined by celestial motions, controlled by gravitation, keep step at all epochs with time as defined by atomic vibrations, controlled by electric forees within the atom? It was suggested in 1938 by P. A. M. Dirac 1 that the constant of universal gravita- tion might be slowly changing with time — with a "time constant" on the order of the age of the universe itself, i.e., about 10 10 years. If this were true, our astronomical and atomic standards would, in the long run, be found to reveal diserepancies. This matter of units and standards is one that most of us do not bother our heads with. We think we know well enough what is meant by a meter and a second; and a ruler or a wateh is usually near at hand. But perhaps the above diseussion may help to suggest that the detailed story of how these basic mea- sures are defined, redefined, and made more and more precise is a quitc faseinating business — especially, perhaps, for time, to ■Dirac, a British theorelical physicist, was one of the leaders in the develop- ment of quantum thcory around 1926-1930. He was awarded the Nobel prize for this work. 65 Units and standards of length and time which astronomical observations over the centuries have con- tributed data of a refinement that almost passes belief. ' SPACE-TIME GRAPHS The primary data in the description of any motion will be a set of associated measures of position and time. Such data might, for example, be a tabulation i n an astronomer's logbook, or a single stroboscopic photograph such as Fig. 2—1. In general the statement of position at any instant will require the use of three coordinates, corresponding to the three independcnt di- mensions of space. I n many circumstances, however, the motion may be confined to a planc, requiring two coordinates only, or to a single line, so that a single positional coordinate sumces. In this last case, and especially if the motion is along a straight line, it is often extremely convenient to display the motion in Fig. 2-13 Example of a space-lime graph for a one-dimen- sional motion. (From the PSSCfilm, "Straight Line Kine- matics," by E. M. Hafner, Edueation Deiselopment Center Film Studio, Newton, Mass., 1959.) Time 'For further reading see, for instance, An Introduction to the Physics of Mass, Length and Time, by N. Fealher, Edinburgh University Press, Edinburgh, 1959. On the matler of time in parlicular, which probably holds the greatest interest, the following books and articles are recommended: J. T. Fraser (ed.), The Voices ofTime, Gcorgc Braziller, New York, 1966; T. Gold and D. L. Schumacher (eds.), The Nature of Time, Corncll University Press, Ithaca, N. Y., 1967; G. M. Clemence, "Siandards ofTime and Frequency," Science, 123, 567 (1956); Lee Coe, "The Nature of Time," Am. J. Phys., 37, 810 (1969); Richard Schlegel, Time and the Physical World, Dover, New York, 1968. 66 Space, time, and motion VELOCITY terms of a space-time graph in which, as a rule, thc time is re- garded as an independent variable, plotted along the abscissa, and the position is plotted along the ordinate. Figurc 2-13 shows such a graph. It has the great merit that it conveys directly, in a way that a numerical table cannot, a complete picture of a given motion. One can immcdiately identify points of maximum and minimum distance from the origin, regions of time in which the motion temporarily ccases altogether, and so on. ' The central concept in the quantitative description of motion is that of velocity. It is a vcctor. The quantitative measure of velocity, in the case of one-dimensional motion, is one of the first pieces of information that we can extract from a space-time graph such as Fig. 2-13. Our way of designating velocities — miles per hour, meters per second, and so on — is a constant reminder of thc fact that velocity is a deriued quantity, based on these separatc mcasures of spacc and time. Has it ever struck you that although physicists have invcntcd names for the units of measurcment of all sorts of physical quantitics, they have never introduced special names for units of velocity? (Seafaring men have done it, though, with their unit the knot, equal to 1 nautical mile per hour.) In nature itself, things seem to be quite otherwise, for although we have not as yet idcntified anything that is directly rccognizablc as a fundamental natural unit of length or time, we do find a fundamental unit of velocity — thc magnitude (c) of the velocity of light in empty space: c = (2.997925 ± 0.000001) X 10 8 m/sec It has become customary in high-energy partiele physics to express velocities as fraetions of c. And in a comparable way, in connection with high-speed (light, the Mach number is used to express the speed of an aireraft as a fraelion or multiple of the speed of sound in air. But this does not alter the fact that 'Such graphical representations of correlated quantities often provide a much more immediale and vivid insight into a situalion ihan do numerical tab- ulations or algebraic formulas, and cven a rough graph, sketehed freehand, can be a great aid to thinking aboul a situalion. A facility in drawing and interpreling graphs as cxpressions of physical relalionships is well worth devcloping. 67 Velocity our basic description of velocities is in terms of the number of units of distance per unit of time. The measurement of a velocity requires at least two measure- ments of the position of an object and the two corresponding measurements of timc. Let us denote these measurements by (ri, Ij) and (r 2 , t 2 )- Using these we can deduce the magnitude and direction of what we can loosely call the average velocity between those points: _ T2—U Vav — t% — /l However, this average velocity is not, in most cases, a very interesting quantity. Sometimes we may find that a graph of s versus / (let us assume a onc-dimensional motion) is a straight linc, so that the value of v deduced from any two pairs of values s and t is the same. But there is a much more basic and general problem: What can we do about defining and evaluating the velocity at an arbitrary instant in a nonuniform motion such as that represented by Fig. 2-13? The next section is devoted to this question, which is of fundamental importance to the whole of the mathematical analysis of motion. INSTANTANEOUS VELOCITY Richard P. Feynman tells the story of the lady who is caught for speeding at 60 miles per hour and says to the police ofncer: "That's impossible, sir, I was traveling for only seven minutes." 1 The lady's objection does not convince us (or the police omcer); we understand that what is at issue is not the persistence of a uniform motion for a long time but the property of the motion as measured over a time interval that might be arbitrarily short. In order to talk about this in specific terms, imagine that along- side a straight section of a road in each direction from a chosen point, P, we have placed a set of equally spaced poles, say at 'This story, further embellished by Feynman with entertaining and instructive details, can be found in The Feynman Lectures on Physics, Vol. I (R. P. Feynman, R. B. Leighton, and M. Sands, eds.), Addison-Wesley, Reading, Mass., 1963. These lectures, of wonderful freshness and originality, range over the whole of physics and provide rich and exciting fare for everyone, whether a beginner or a veteran in the subject. Feynman, one of the most outstanding physicists of our time, was awarded the Nobel prize in 1966 for his fundamental contributions to quantum field theory. 68 Space, time, and motion Fig. 2-14 Arrange- menl for inferring the instantaneous velocity of a car as it passes the point P. 5-meter intervals (see Fig. 2-14). On each pole is an electric clock with a sweep second hand. The clocks are synchronized and run continuously. A movie camera is placed opposite each pole, so as to photograph the pole, the clock face, and the road. A car comes along the road; the cameras are set running and the photographic records are developed. It is like the photofinish of a race. Each film will contain one frame in which, let us say, the front bumper of the car is more or less exactly in line with the pole. We now assemble the records. Take the data for the poles in pairs, equal distances before and after the central point P. For a given pair let the separation in distance be called As (a multiple of 10 m) and let the differcnce of time readings be called A;. Then the ratio As/A/ is the average velocity over the range of distance Ax centered on P. Now construct a graph, as shown in Fig. 2-1 5(a), of As/A/ as a funetion of As. Unless the motion of the car has been extraordinarily erratic, the points can be fitted by a smooth curve that flattens out for the smaller values of As. We extrapolate it baekward, and the value of As/A/ for As = is our measure of the instantaneous velocity at the central point P. We could equally well plot the values of As/A/ against the time intervals A/, as shown in Fig. 2- 15 (b); the value of As/A/ at A/ = is again the same, even though the Fig. 2-15 (a) Eual- ualion ofthe instan- taneous velocity ds/dt by extrapolation to As m of a graph o f As/At against As. {b) Evalualion of the instantaneous velocity ds/dt by extrapola- tion WAt = of a graph ofAs/At against At. 69 Instantaneous velocity Fig. 2-16 The "slope" of a graph ofone physica! guantity against another is not primarily geometrical; U is defined by the ratio of the changes A Q and Aq, each measured in whatever unils are appropriale. graph itself looks somewhat different. And here, in physical terms, is what we mean by evaluating the limit of the average velocity, when the range of position or time over which As/At is evaluated is shrunk to zero. The process described above corresponds to the mathe- matical process of determining a derivative — in this case, of displacement with respect to time. Using the Standard calculus notation, which in the context of the above discussion almost speaks for itself, we write instantaneous velocity o = lim — = — A«-oA/ dt (2-5) The quantity ds/dl, a synonym for the limit of As/At, is, in mathe- matical parlance, the first derivative of 5 with respect to t. In geometrical terms, it reprcsents the slope of (a tangent to) the graph of s versus t at a particular value of t. [A note concerning the mcaning of the word "slope" in graphs of physical data is in order. The graph — e. g., as in Fig. 2-16 — is a display of the numerical measure of one physical quantity plotted against the numerical measure of another, using scales that are entirely arbitrary and dictated by convenience alone. Thus there is, in general, no physical significance to the inclination of such a line as measured by a protractor. By "slope" we mean simply the ratio of the change in the quantity repre- sented on the ordinate to the corresponding change in the quan- tity represented on the abscissa— e.g., (Q 2 — <2i)/(<72 — <7i)- In this ratio, numerator and denominator are each a pure number times a unit, and the slope is then expressible as the quotient of these numbers, labeled with its own characteristic units — e.g., metcrs per second.] If the same ideas as above are applied to changes with time of the displacement in space, we arrive at a general definition of instantaneous velocity as a veetor: .. Ar dt v = hm — = — Ar—O Ar dt (2-6) In Fig. 2-17(a) we indicate what the evaluation of Ar entails. Clearly, if wc display the path of an object as a continuous curve, as in Fig. 2-17(b), the instantaneous velocity veetor is tangent 70 Space, time, and motion Fig. 2-17 (a) Vector diagram to define a small chaiige ofposi- tion. (b) Instan- taneous velocity vec- tor, tangent to the path. to this curvc; this is implicit in the definition. With this departurc from straight-line notion, we shall draw attention to the distinetion that is made technically (but not always faithfully observed) bctwcen the words velocity and speed. The speed is the magnitude of the vector velocity v. The spced is thus, by definition, a positive, scalar quantity. The o that figures in our analysis of straight-line motion has, in faet, represented the velocity; it may take on negative as well as positive values, which of course is all the information needed to specify the direetion of motion in the one-dimensional case. Even if we are dealing with motion along a single straight line, the use of the vectorial deseription of position and velocity will be necessary if displacements are referred to an origin not Fig. 2-18 (a) Vector positions in a straight-line motion, referred to an origin not on the line. (b) Corresponding diagram for a curvilinear path. 71 Instantancous velocity on the line itself. It is, in fact, a very instructive thing to consider a straight-line motion from this point of view [see Fig. 2-18(a)], because it helps us to scc the one-dimensional situation in its larger context. We can become accustomed to the idea that the displacement vcctor Ar is, in general, in a different direction from the position vcctor r, when the latter is referred to an arbitrary origin O. This then makcs the transition from the description of a rectilinear path to the description of an arbitrary curved path [Fig. 2-18(b)] scem less abrupt. It also emphasizes the fact that although it may be very convcnient, in the case of straight-line motion, to choose an origin on the line itself, this is certainly not necessary and may not always be possible. RELATIVE VELOCITY AND RELATIVE MOTION Sincc the vector velocity is the time derivative of the vector dis- placement, the velocity of one object relative to another is just the vector difference of the individual velocities. Thus if one object is at r! and another object is at r 2 , the vector distance R from object 1 to object 2 is given by R = r2 - n The rate of changc of R is then the velocity, V, of object 2 relative to object 1 , and wc have V m — - — — — dt dt dt i.e., V = v 2 - vi ( 2 ~ 7 ) This relative velocity V is the velocity of object 2 in a frame of reference attached to object 1. In diseussing frames of reference earlier in this chapter, we pointed out how the choice of some particular frame of reference may be advantageous because it gives us the clearest picture of what is going on. Nothing could illustrate this better than the practical problems of navigation and the avoidance of collisions at sea or in the air. Imagine, for example, two ships that at some instant arc in the situation shown in Fig. 2-19(a). The veetors v, and v 2 represent their velocities (which we take to be constant) with respect to the body of water in which they both 72 Space, time, and motion Fig.2-19 (a)Paths of two ships moving at consiam velocity along courses that intersect. (b) Path of ship B relative to ship A, showing that they do not collide even lhough their paths cross. move. The paths of the ships, extended along the directions of motion from the initial points A and B, intersect at a point P. Will the ships collide, or will they pass one another at a safe distance? The answer to this question is not at all elear if we stiek to the ocean frame, but if we deseribe things from the standpoint of one of the two ships the analysis becomes very straightforward. Let us imagine that we are standing on the deck of the ship marked A . Putting ourselves in that frame of reference means giving ourselves the velocity Vi with respect to the water. But from our standpoint it is as if the water, and everything else, were given a velocity equal and opposite to vj. Thus to every motion as observed in the ocean frame we add the veetor — Vi, as implied by Eq. (2-7). This automatically, and by definition, brings A to rest, as it were, and shows us that the velocity of the ship B, relative to A, is obtained by combining the veetors v 2 and —V], as shown in Fig. 2-1 9(b). The veetor distance between the ships is unaffected by this change of viewpoint. So now we can see the whole picture. B follows the straight line shown, as indicated by several successive positions in the diagram. It will miss A by the distance AN, the perpendicular distance from A to the line of V. The time at which this elosest approach oecurs is equal to the distance BN divided by the magnitude of V. Thus B seems to sweep aeross /4 's bow, more or less sideways. If you have had oceasion to observe a elose encounter of this sort, especially if it is out on the open water with no landmarks in sight, you will know that it can be a curious experience, quite 73 Relative velocity and relative motion disturbing to the intuitions, because the observed motion of the other ship seems to be unrelated to the direction in which it is pointing. PLANETARY MOTIONS: PTOLEMY VERSUS COPERNICUS Some of the most fascinating problems in the study of motion, and in particular of its relative character, have arisen in man's attempts to elucidate the motions of the heavenly bodies, in- cluding our own earth, through space. Observational astronomy, the first of the exact sciences, has yielded data of marvelous accuracy for several thousand years. But the question has always been how to interpret these data. Let us consider some of the main features of the problem. The first thing to recognize is that naked-eye astronomy is, Fig. 2-20 (a) Palh of Venus omong the stars during a 6- month period, showing reoersed (retrograde) motion at one stage. (b) Similar set ofob- servalions on Mars. {.Bot h diagrams after E. M. Rogers, Physics for the In- quiring Mind, Princeton Uniuersity Press, Princeton, N. J., 1960.) IA Space, time, and motion almost exclusively, the study of directions rather than distancei These unaided observations reveal nothing about the distances of the stars. The great Greek astronomers Aristarchus (third century B.c.) and Hipparchus (~150b.c.) did make reasoned estimates of the distances of sun and moon (the latter very successfully), ' but the only direct clues to the distances of the planets are through such quantitative evidence as changes of apparent brightness with time, suggesting that whatever the distances of the planets from the earth may be, they undergo systematic variations. Thus, as is still the practice, the positions of astronomical objects are defined in the first place in terms of their directions only (this being all we need to find them with a telescope) and can be described as if they were points on the surface of a sphere of large but arbitrary radius — the celestial sphere — with its center at the earth, and with a polar axis and an equator defined by the earth's own axis of rotation (cf. Fig. 2-2). In these terms the primary data on the motions of the planets are of the type shown in Fig. 2-20. It is already a tribute to the genius of the early astronomers that they were able to visualize such strange-looking paths as the projections, on the celestial sphere, of orbital motions of various kinds. In particular, the belief took hold that the orbits must be combinations of circular motions. This is not the place to go into a detailed account of the problem; some outstandingly fine accounts exist elsewhere. 2 Instead, we shall simply focuson an idealized presentation of the two main models: an earth- centered (geocentric) or a sun-centered (heliocentric) solar system. The most intuitively reasonable picture of the universe, in terms of everyday experience, is undoubtedly one that places the earth at the center of everything. Not one of us, without benefit of hindsight, could interpret his first impressions in any other way, and the ancient descriptions, such as the biblical one in Genesis, are entirely justifiable in these terms. It was the Alexandrian astronomer Ptolemy (—150 a.d.) who built this picture into a quantitative model of planetary motions and 'See the problems at the end of this chapter, and also Chapter 8. 2 See, for example, E. M. Rogers, Phyiics for the Inguiring Mind, Princeton University Press, Princeton, N.J., 1960, Chaps. 12-18, or E. C. Kemble, Physical Science, lis Structure and Developmenl, MIT Press, Cambridge, Mass., 1966, Chaps. 1-5; or any of a number of excellent books on elementary astronomy. 75 Planetary motions: Ptolemy versus Copernicus Fig. 2-21 (a) Apparent matian of a planet as explained in the Ptolemaic syslem. The planet P moues on the epicycle whose center C follows a circular path around the earth, E. {b) Copernican explanation ofobserved motion. The epicycle of (a) Is seen as being a reflection of the earth' s own motion around the sun. described it in his great work, the Almagest. Figure 2-21 (a) illustrates the essential features. Setting aside the effects of the earth's daily rotation, the motion of the sun is an approximately circular path, with a period of 1 year, around the earth, E, as center. l The motion of a planet, however, is compound. It can be fairly closely approximated by imagining that a point C travels uniformly around a circular path, and that the planet, P, travels in another circular path with respect to C as center. This extra circle is called an epicycle; the combination of these two motions, if they are in the same plane, gives rise to a complicated path that can have backward loops as shown. Precisely the same result would be obtained if we interchanged the roles of the two circles. (Verify this.) If we now imagine viewing this motion from the earth, and projecting it onto the celestial sphere or to any other constant distance, we obtain almost the kind of variation of angular posi- tion that is shown in Fig. 2-20. We can come even closer by tilting the plane of the epicycle out of the plane of the primary ■This path, which carries the sun eastward through the constellations around the celestial sphere, is known as the ecliptic. 76 Space, time, and motion circle a little. To fit the motion of a particular planet it is neces- sary to choose appropriate values for the ratio of the radii of the two circles and also for the times to make one complete circuit of each. It is a noteworthy fact that in two cases (Mercury and Venus) the period of the primary circle is exactly 1 sidereal year (i.e., the period for one complete orbit of the sun around the ecliptic), and in the other three cases (Mars, Jupiter, and Saturn) the period of the epicycle is 1 sidereal year. Indeed, this can be taken as the crucial clue to what we now regard as the truer picture. Suppose that we now place the sun at the center, as in Fig. 2-21 (b), and make the earth travel around a circle that has the same radius as the epicycle in Fig. 2-21 (a). Then if the planet, P, travels around another simple circle, of radius equal to that of the primary circle in Fig. 2-21 (a), the relative positions and velocities of E and P can be made precisely the same as before. We have drawn the positions of E and P in the two diagrams to display this exact correspondence. The reason for the appearance of the sidereal year in one or other of the two component motions of a planet in the Ptolemaic model is now very clear, and much of the arbitrariness of the whole description disappears. It is this new, heliocentric, description with which we associate the name of Copernicus. He presented it in great detail in his principal work, De Revolutionibus Orbiwn Celestium ("On the Revolutions of the Celestial Spheres") published in 1543, the year of his death. Actually, the first suggestion that the sun occupied the central position had been made about 1800 years earlier by Aristarchus, who from other observations knew that the sun was much bigger than the earth (although he seri- ously underestimated just how huge and distant it is). There is no record, however, that the heliocentric theory was developed in quantitative detail before Copernicus. It is interesting, by the way, to note his clear understanding of the relativity of motion. Here is a translation of his own statement of the principle: "For all change in position which is seen is due to a motion either of the observer or of the thing looked at, or to changes in the position of both, provided that these are different. For when things are moved equally relatively to the same things, no motion is perceived, as between the object seen and the observer." l 'From Arthur Berry, A Short History of Astronomy, Dover Publications, New York, 1961. 77 Planetary motions: Ptolemy versus Copernicus PROBLEMS Given the persistent intrusion of the sidereal year into the Ptolemaic scheme, it may seem surprising that the heliocentric picture did not prevail at a much earlier stage, especially when its possibility had been recognized by Aristarchus about 400 years before Ptolemy's day. It must be remembered, however, that we have been presenting a greatly oversimplified model of the solar system, and the ancient astronomers were legitimately worried over discrepancies between thcse idcalized models and the precise, hard facts of observation. Both Ptolcmy and Coper- nicus were driven to introduce numerous auxiliary circular motions to obtain even approximate agreement between theory and observation. The Copernican scheme, in the form in which Copernicus himself developed it, was not in fact notably less arbitrary or less complex than the Ptolemaic. Not until the theory could free itself of the cirele as the basis of all celestial motion was a fundamental solution to be finally attained, al- though the introduetion of dynamical considerations — the laws of foree and the laws of motion — transformed the context within which the observations were interpreted. We shall come back to these questions in Chaptcr 8 and, more fully, in Chaptcr 13. 2-1 Slarting from a point that can be laken as the origin, a. ship travels 30 miles northeast in a straight line, and then 40 miles on a course that heads SSW (a direetion making a countcrclockwise angle of 247^° with a reference linc drawn eastward). Find the x and y coordinates of its final posilion (x eastward, y northward) and its distance from the slarting point. 2-2 The scalar (dot) produet of two veetors, A - B, is equal to ABcosOah, where 6 mi is the angle between the veetors. (a) By expressing the veetors in terms of their Cartesian compo- nents, show that A Z B X + AyBy + A,B Z cos d a u = AB (b) By using the relation between rectangular and spherical polar coordinates [Eq. (2-4)], show that the angle 0i2 between the radii to two points (R, 6\, <p\) and (R, 02, <P2) on a sphere is given by cos0i2 = cos0i cos 62 + sin 0i sin $2 cos (^2 — <pi) (Notc that the distance between the two points as measured along the 78 Space, timc, and motion great circle that passes through them is equal to Rd 12, where 0i2 is expressed in radians. This can be used, for example, to calculate mileages between points on the earth's surface.) 2-3 (a) Calculate the Cartesian coordinates of New York, U.S.A. (41° N, 74° W) and Sydney, Australia (34° S, 151° E). Take an origin of coordinates at the earth's center, with a z axis through the north pole and an x axis passing through the equator at the zero of longitude. The earth's mean radius is 6370 km. (b) Find the distance along an imaginary straight tunnel bored through the earth between New York and Sydney. (c) Compare the result of (b) with the shortest practicable route between these points by a great-circle flight. You can either calculate this, using the result of Problem 2-2(b), or measure it directly on a globe with the help of a piece of string. 2-4 (a) Starting from a point on the cquator of a sphere of radius R, a partiele travels through an angle a eastward and then through an angle /3 along a great circle toward the north pole. If the initial position of the point is taken to correspond to x — R, y = 0, z = 0, show that its final coordinates are R cos a cos /3, R sin a cos /3, and R sin 0. Verify that x 2 + y 2 + z 2 = R 2 - (b) Find the coordinates of the final position of the same par- tiele if it first travels through an angle a northward, then changes course by 90° and travels through an angle /3 along a great circle that starts out eastward. (c) Show that the straight-line distance As between the end points of the displacements in (a) and (b) is given by As 2 = 2« 2 (sin/3 - sin a cos P) 2 (d) Using the above result, check the statement in the text (p. 60) that there is a difference of about 40 miles in the end points of (1) a displacement of 1000 miles eastward on the earth along the equator, followed by a displacement 1000 miles north, and (2) a dis- placement of 1000 miles north, followed by a displacement of 1000 miles starting out eastward on another great circle. (Put Ra = Rp = 1000 miles. The approximation cos 6 s» 1 — d 2 /2 will be found useful.) 2-5 If you found yoursclf transported to an unfamiliar planet, what methods could you suggest (a) To verify that the planet is spherical? (b) To find the value of its radius? 2-6 The radius of the earth was found more than 2000 years ago by Eratosthenes through a brilliant piece of analysis. He lived at Alex- andria, at the mouth of the Nile, and observed that on midsummer day at noon, the sun's rays were at 7.2° to the vertical (see the figure). He also knew that the people living at a placc 500 miles south of 79 Problcms Equator / / / 500 ^-AleKandria 7.2° ^ milesi\ sun Alexandria saw the sun as being directly overhead at the same date and time. From this information, Eratosthenes deduced the value of the earth's radius. What was his answer? 2-7 It has been suggested that a fundamental unit of length is repre- sented by a distance about equal to a nucleon diameter, and that a fundamental unit of time is represented by the time it would take a light signal (i.e., the fastest kind of signal achievable) to travel across a nucleon diameter. Express the radius of the universe and the age of the universe in terms of these units, and ponder the results. 2-8 A particle is confined to motion along the x axis between re- fiecting walls at x = and x = a. Between these two limits it moves freely at constant velocity. Construct a space-time graph of its motion (a) If the walls are perfectly reflecting, so that upon reaching either wali the particle's velocity changes sign but not magnitude. (b) If upon each reflection the magnitude of the velocity is reduced by a factor/(i.e., i>2 = —foi). 2-9 A particle that starts at x = at t = with velocity +v (along x) collides with an identical particle that starts at x = xo at / = with velocity —v/2. Construct a space-time graph of the motion before and after collision (a) For the case that the particles collide elastically, exchanging velocities. (b) For the case that the particles stick together upon impact. 2-10 A particle moves along the curve y = Ax 2 such that its x posi- tion is given by x = Bt. (a) Express the vector position of the particle in the form r(f) = xi + yy (b) Calculate the speed v (= ds/di) of the particle along this path at an arbitrary instant /. 2-11 The refraction of light may be understood by purely kinematic considerations. We need to assume that light takes the shortest 80 Space, time, and motion (timewise) path between two points (Fermat's principle of least time). Referring to the figure, let the speed of light in medium 1 be vi and in medium 2, i?2- Calculate the time it takes light to go from point A to point B as a function of the variable x. Minimize with respect to x. Given that vi = c/rti and V2 = c/ri2 where the n's are known as indices of refraction, prove Snell's law of refraction: «i sinfli = /»2sin02 2-12 At 12:00 hours ship A is 10 km east and 20 km north of a certain port. It is steaming at 40 km/hr in a direction 30° east of north. At the same time ship B is 50 km east and 40 km north of the port, and is steaming at 20 km/hr in a direction 30° west of north. (a) Draw a diagram of this situation, and find the velocity of B relative to A. (b) If the ships continue to move with the above velocities, what is their closest distance to one another and when does it occur ? 2-13 The distance from A to B is /. A plane fiies a straight course from A to B and back again with a constant speed V relative to the air. Calculate the total time taken for this round trip if a wind of speed v is blowing in the following directions: (a) Along the line from A to B. (b) Perpendicular to this line. (c) At an angle d to this line. Show that the time of the round trip is always increased by the existence of the wind. 2-14 A ship is steaming parallel to a straight coastline, distance D offshore, at speed V. A coastguard cutter, whose speed is v ( < V) sets out from a port to intcrcept the ship. 81 Problems q^* -f: -o= *£ Port Bulge rotates to keep pace with moon- ^Earth's daily rotation (a) Show that the cutter must start out before the ship passes a point a distance D(V 2 — v 2 ) 1 ' 2 1 o back along the coast. (H ini: Draw a vector diagram to show the vclocity of the cutter as seen from the ship.) (b) Tf the cutter starts out at the latest possible moment, where and when does it reach the ship ? 2-15 With respect to the "fixed stars/' the earth rotates once on its axis in one sidereal day — that is how the sidereal day is defined. (a) The length of the year is about 366 sidereal days. By what amount is the mean solar day (from noon to noon) longer than the sidereal day ? (b) The moon completes one orbit with respect to the stars in 27.3 sidereal days. That is, in this time the line from earth to moon turns through 360° with respect to the stars. The time between cor- responding high tides on successive days is longer than 1 solar day (24 hr) because of this motion of the moon. (The high tide is an ocean bulge at a fixed direetion with respect to the moon— see the figure.) Show that the daily lag is elose to 50.5 min (60 min = ^ solar day). 2-16 (a) The orbital radii of Venus and Mars are 0.72 and 1.52 times the radius of the earth 's orbit. Their periods are about 0.62 and 1 .88 times the earth's year. Using these data, construct diagrams by which to find how the apparent angular positions of Venus and Mars change with time as seen from the earth, assuming that the orbits of all three planets lie in the same plane. Compare your results with Fig. 2-20. (b) With respect to the ecliptic (the plane of the earth's own orbit) the planes of the orbits of Venus and Mars are tilted by about 3.5° and 2°, respectively. Consider how the apparent paths of Venus and Mars are affected by this additional feature. 2-17 (a) What methods can you suggest for finding the distance from the earth to the moon (without using radar or space fiight) ? Moon 82 Space, time, and motion Moon Observer on earth - (b) The astronomer Hipparchus, more than 2000 years ago, found the distance of the moon as a multiple of the earth's radius by observing the duration of a total eclipse of the moon by the earth (see the figure). The rays from the sun have a spread of directions of about ±5°, and the moon itself subtends an angle of just about ^° (the same as the sun within about 2%). The moon takes about 29 days to circle the earth, and the duration of the total eclipse is about l£ hr. Use these data to obtain the moon's distance. 2-18 The astronomer Aristarchus had the idea of comparing the distances of the sun and the moon from the earth by measuring the angular separation 6 between them when the moon was exactly half full (see the figure). Using our present knowledge of what these distances are, criticize the feasibility of the method. Aristarchus found 6 = 87°. What result would this imply? Calculate what the angle really is and what error would be introduced in the distance if this angle were uncertain by ±0.1°. 83 Problems At present it is the purpose of our Author merely to investigate and to demonstrate some of the properties of accelerated motion (whatever the cause of this acceleration may be). galileo, Dialogues Concerning Two New Sciences (1638) Accelerated motions ACCELERATION from THn purely descriptive point of view, the central feature of motion is velocity — the instantaneous rate of change of position with time. But we must dig a little deeper to get to the quantity that proves to be the crucial one in relating motion per se (kinematics) to motion as governed by forces (dynamics). This is acceleration, the rate of change of velocity with time. Again we shall develop the basic ideas in the first instance in the context of straight-line motion. If the instantaneous velocity is found to be a linear function of time, as in Fig. 3-1 (a), then we conclude that the instantaneous acceleration is the same at all times and equal to the slope of the graph as evaluated from any two points on it. ! If, on the other hand, the values of the instan- taneous velocity define some sort of curve, as in Fig. 3-1 (b), then we must obtain the instantaneous acceleration at a par- ticular value of f by a limiting process: • • i- Ad dv .. ,. instantaneous acceleration a = lim — — — (3-1) a,_o A? di Thus the acceleration is the first derivative of the velocity with 'The units in which this is expressed will be defined by our chosen unit of velocity divided by whatever unit of time is found convenient. For example, the acceleration of a car is expressed most effectively and vividly in m.p.h. per second. But in the more analytical treatment of motions it is almost essential to use the same time units throughout, e.g., expressing velocities in m/sec and accelerations in m/sec 2 . Otherwise, when we see a symbol such as i/, we have to stop and ask ourselves which of the different units it is measured in, and that makes for confusion and error. 85 Fig. 3-1 (a) Oraph ofv versus t for a uniformly acceleraled motion. (b) Motion with varying accelera- tion. In the indicated time interval Al ihe acceleralion is negative. v(/ + A/) respect to time. If, however, we wish to tie our definition of acceleration to a priinary record of position against time, then wc can write it as the second derivative of s with respect to /: (3-2) In general, we must be ready to take into account a variation of velocity in direction as well as magnitude. This then requircs us to considcr the acceleration explicitly as a vector quantity. Just as we previously considered vectorial changes of position, so now we can show the instantaneous velocity vectors at two neighboring instants, as in Fig. 3-2, and can proceed to a state- ment of the instantaneous vector acceleration, a: .. Av dv d r '-^oAt-di-dfi As far as kinematics by itself is concerned, there is no good reason why we should stop here. We could define and evaluate the rate of change of acceleration, but in general this does not represent information of any basic physical interest, and so our discussion of mechanics is based almost exclusively on the three quantities displacement, velocity, and acceleration. You may fcel, cspccially if you have some prior familiarity with calculus, that we have gone to excessive lengths in our dis- Fig. 3-2 Small change of a velocity that is changing in bot/i magnitude and direction. 86 Accclcratcd motions cussion of instantaneous velocity and acceleration. But make no mistake about it; these are very subtle concepts. The notion that an object could both be at a certain point and moving past that point was one that perplexed some of the best minds of antiquity. Indeed, it is the subject of one of the famous paradoxes of the Greek philosopher, Zeno, who contended that if an object was moving it could not be said to be anywhere. l If you want to test your own mastery of these ideas, try explaining to someone how an object that is at a certain point with zero velocity (i. e., instantaneously at rest) can nonetheless move away from that point by virtue of having an acceleration. It really isn't trivial. THE ANALYSIS OF STRAIGHT-LINE MOTION Given a detailed record of position versus time in a straight-line motion, the procedures that we have described enable us to find the associated variations of velocity and acceleration. The sequence of diagrams in Fig. 3-3, going from top to bottom, shows an example of this. But how about the converse of this process: given the acceleration as a function of time, to infer the graphs of velocity and displacement? The basic definitions of velocity and acceleration suggest the appropriate procedure for doing this. From Eq. (3-1) we see that the change of velocity Av in a short time At is given, at least approximately, by the equation Av = aAt Of course, "approximately" is not good enough, but we recognize that the smaller we choose At to be, the more nearly is Av an accurate statement of the change of v. Again we resort to a graphical presentation. In Fig. 3-4(a) we show a graph of ac- celeration versus time ; it then becomes apparent that a At can be read as the area of a narrow rectangular strip, the top of which cuts aeross the curve of a versus t. From there it is a short step to concluding that the over-all change of v between two given values of t is obtained by summing all such rectangular con- tributions. (We must, of course, recognize that wherever a is negative, the area represented by a At, and hence the change of v, is also negative.) We then imagine that the widths At are made 'On a quite different basis, the uncertainty principle of quanlum mechanies expresses our inability to measure both the position and the velocity of a moving object with arbitrarily high precision. 87 The analysis of straight-line motion Fig. 3-3 Sel ofre- lated graphs showing the time dependence of (a) posilion, (b) velocity, and (c) acceleralion. (a) vanishingly small so that the sum of all the strips coincides, in this limit, with the area under the smooth curve of a versus t. This is then written mathematically as a definite integral: u 2 — 01 = aU) di (3-3) wherc we write a(t) to show that the acceleration is to be con- sidered as a specific function of time. Most often this integral is evaluated up to some indefinite time /, starting from some choscn zero of time at which the velocity is v . Thus we put — 00 = fl(0 di (3-4) Notice, then, that our integral, starting from knowledge of the acceleration as a function of time, gives us only the change of velocity during the time /. Information about the value of v at t = (or at some other specific time) must be supplied sep- 88 Accelcratcd motions (a) (b) Fig. 3-4 (a) Graphical integration o/a graph of acceleration versus time to find change ofvelocity. (b) Graphical integration of velocity-time graph to find displacement. arately; v is a typical example of a constant of integration that requires some knowledge of the initial conditions — or of the specific value of v at any one value of t. In like manner, given the curve such as that of Fig. 3-4(b) of v against / (which may represent the initial data or may itself have come from the above integration of the acceleration function) we can proceed to find the distance traveled. It is represented by the area contained between the velocity-time curve, the r axis, and the ordinates at two given values of t: S2 - si = v(t) dt (3-5) Again it is most usual to evaluate the integral from / = up to an arbitrary time, and again a constant of integration — the position s o at t = — must be supplied : s — Si) = v(t) dt (3-6) Very often, of course, it is possible to choose Sq = 0, but one should never forget that the area included under the velocity- time graph gives us only the change of position. The simplest applications of these kinematic equations, for a = or a = constant, are undoubtedly familiar to you. In the first case the velocity-time graph is simply a rectangle, as shown in Fig. 3-5(a). If the acceleration is constant (but not zero), 89 The analysis of straight-line motion Fig. 3-5 (a) Veloc- ity-time graph for the special case ofzero acceleration. (b) Velocity-lime graph for a constant (positice) accelera- tion. Fig. 3-5(b) is appropriate. The magnitude of the constant ac- celeration is given by a = slope = v — do This conforms to the definition of acceleration in Eq. (3-1). If we took a as given, then we would obtain this same result by integration, according to Eq. (3-4), V — DO i / dt - Jo at (3-7) The area in Fig. 3-5(b) that represents the distance traveled can be thought of as made up of the two shaded regions as shown. Hence s — so = vot + \{o — Uo)l Combining the last two equations, we get s — so = Dot + \at 2 which we can recognize also as the result of evaluating the integral in Eq. (3-6) with v(l) = v + at: — so = / ./o (dq + at) dt = Dot + %at (3-8) It is sometimes convenient to remove all explicit reference to the time, by combining Eqs. (3-7) and (3-8). This gives us o 2 = u 2 + 2a(s - so) (3-9) 90 Accelerated motions For ease of reference, we repeat these equations below as a group: Kinematic ( o = v + at (3-10a) equations l v 2 = vq 2 + 2a(s — so) (3-10b) (valid only for constant a){ s = s + o t + \at 2 (3-1 Oc) Although these mathematical expressions for accelerated motion are tidy and extremely useful, it should be remembered that a truly constant acceleration is never maintained indefinitely. For example, the problems that everyone learns to solve on free fail under gravity, using a constant acceleration g, really do not correspond to the facts, because air resistance causes the ac- celeration to become less as the velocity increases. For low velocities the error may not be big enough to worry about, but it is there. Later we shall be dealing with situations in which the acceleration varies in some mathematically well-defined way with position or time. Thus the emphasis will shift away from Eq. (3-10) and toward the more general statements expressed in Eqs. (3-4) and (3-6). Another very important factor in solving real problems in kinematics is the digital computer. Whether or not the acceleration i s described by a mathematically convenient function, the actual technique of getting numerical answers to problems on motion — e.g., the path of a rocket or a satellite — will be the summation of small but finite contributions, corre- sponding to the strips of Fig. 3-4. The program for solving problems in motion is then represented not by mathematical integrals, but by equations such as the following: v(t +&t) = v(t) +a(/)Ar s(/+A/) = s(l) + v(t)At s(t) = so +!>(') A' There are many problems in motion that can be handled as one-dimensional problems, even though the space in which dynamical processes go on is a space of three dimensions. A prime reason for this is that it is often feasible to resolve the vectors of position, velocity and acceleration into their com- ponents in a rectangular coordinate system and then proceed to work with the separate components. Under these conditions, as we have mentioned before, it is not necessary to make use of vector notation as such, even though we know that we are dealing 91 The anaiysis of straight-line motion with directional quantities. It becomes sufficient to choose an axis of reference along the given line of the motion and adopt a convention that selects one direction along this axis as positive and the opposite direction as negative. Which we choose as positive is arbitrary, but having made a choice for the purpose of a particular problem we must stick to it. Directed quantities that are to be found as a result of a calculation — e.g., the un- known final coordinate, x, of a particle moving along a straight line — will always be taken as measured along the positive direc- tion of the axis. If the answer comes out negative, this auto- matically tells us that the final position is on the negative side of the origin. In other words, it is not necessary (and may be actually inadvisable, because it can lead to confusion) to inject preconceived ideas as to which sign or direction a quantity will be found to have; the mathematics will do it for you. Example. A particle starts out at / = from the point x = 10 m with an initial velocity v = 15 m/sec and a constant acceleration a = — 5 m/sec 2 in the x direction. Find its velocity and position at t = 8 sec. We have o(0 = oo + at d(8) = 15 + (-5)(8) = -25 m/sec Also x(t) = x + v l + ial 2 x(8) = 10 + (15)(8) + |(-5)(8) 2 - -30 m Thus at t = 8 sec the particle has passed back through the origin to a point on the far side and is traveling in the direction of in- creasingly negative x. The whole progress of the motion up to t = 8 sec is shown in the two graphs of Fig. 3-6. Notice how much information is provided at a glance in these diagrams. One sees at once how the velocity falls to zero at / = 3 sec and at t = 6 sec becomes equal and opposite to its initial value. One sees how the maximum value of x corresponds to the instant at which u reaches zero and reverses sign, and that the particle returns to its original displacement (x ) at t = 6 sec when v has reached the negative of its initial value v Q , so that the total area under the velocity-time curve up to that instant is zero. Even though this is a very simple and straightforward example, it dis- 92 Acceleratcd motions Fig. 3-6 (a) Veloc- ity-time graph for specific values of ini- tial velocily and con- sianl (negatiue) acceleration. (b) Posiiion versus timefor the motion represented in (a). m 40 30 20 10> /-20 plays many features that are worthy of note. And seeing how the various details fit together will greatly strengthen one's grip of basic kinematics. A COMMENT ON EXTRANEOUS ROOTS Occasionally, in turning the mathematical handle in the solution of the kinematic equations, one cranks out extraneous roots that are contrary to the physical situation. How docs one recognize these "incorrect" answcrs and when can thcy be discarded? Many of thcsc cxtraneous answers have their origin in the fact that in solving a problem we always state initial conditions which specify the situation at the moment we first begin to follow the motion of the particle at t = 0. Specifying the position and velocity at t = does not tell us anything whatsoever about the past dynamical history of the particle. lndeed, it is of no con- sequence whatsoever how the particle attained these initial values. For example, if we think of the motion of a body falling freely from rest at an initial height h above the earth's surface, we have 93 A comment on exlraneous roots v = O and y = h. We may have held the body at its initial position and released it from rest, or we may have thrown it upward so that it rises to a maximum height h. The motion of the particle subsequent to the instant of time (t = 0) when it was at the height h with zero velocity is identical in both cases. The most frequent source of extraneous roots is the equation relating displacement to time in motion with constant acceleration — i.e., Eq. (3-10c). Mathematically, this is a quadratic equation for t and must be solved as such if the displacement is given and the time is to be found. Because initial conditions alone do not give us information about the earlier motion of the particle (unless additional relevant data are given), a root of this equation corresponding to a negative value of t may not be valid. To illustrate this, consider the problem used as an example in the last section. Suppose we ask for the value of / at which x = 0. We have x(t) = 10 + 15? - £/ 2 Putting x = 0, we get the following quadratic equation: i 2 - 6t - 4 = with the roots t = 3 ± V 13 = 6.6 sec or — 0.6 sec. Figure 3-6(b) makes quite apparent the origin of these two roots and shows how the negative root follows from an extrapolation of the graph backward into the region of times prior to t = 0. But that may be quite unjustified. We might, if asked, say: "Oh yes, I held the particle at + 10 m from the origin until t = and then fired it off in the positive x direction with its initial velocity of 15 m/sec." If that were the case, the solution x = Oat* = —0.6 sec would be a complete fabrication; one would be forced to recognize that it simply did not correspond to reality. One should not, however, discard extraneous roots without first asking, in the way we have just done, whether there is a clear physical reason for doing so. The matter of extraneous roots has had an interesting con- sequence in the history of physics. In quantum mechanics, developing a relativistically correct equation led to two values for the total energy of an electron : positive and negative. Nega- tive values were initially rejected outright as having no physical significance. After all, what meaning can one attach to a kinetic energy less than zero? However, at a later time, P. A. M. Dirac investigated more carefully the nature of these negative energy 94 Accelerated motions Fig. 3-7 Ideal ized parabolic trajectory for motion under grav- ity in the absence of air resistance. states and was led to a highly successful theory of electrons which predicted the existence of positrons and other "antimatter" particles. TRAJECTORY PROBLEMS IN TWO DIMENSIONS One of the most famous and widely studied problems in motion is that of free fail near the earth's surface. It provides an illus- tration of the fact that, in the rectangular-coordinate system represented by horizontal and vertical directions, the two ortho- gonal components of the motion are completely independent (provided that air resistance is negligible— see p. 225). The path of an object may be treated as two, separate motions occurring simultaneously, and each may be analyzed as if the other were not present. Galileo, in his Dialogue on the Two Chief World Systems (1632), first recognized this fact. 1 We shall consider the motion of an object hurled with initial speed v at an angle as shown in Fig. 3-7 from a height h 'The accomplishmenls of Galileo Galilei, born in Pisa in 1564, the year of Shakespeare's birth and Michaelangelo's dealh, are often citetl as the be- ginning of modern science. Galileo's publication on astronomy, Dialogue on The Two Chief World Sysiems, incorporated the Copernican model and led to confiicts with church authorities. While technically a prisoner of the Inquisition, Galileo turncd to the studies of mechanics and published (1638) surreptitiously in Holland the results of his investigations "Discourses and Mathematical Demonsirations Concerning Two New Sciences Pertaining to Mechanics and Local Motion," commonly referred to as Two New Sciences (translated, Dover Publications, New York). These books, written largely in the form of imaginary conversations, have a surprisingly modern flavor, and impress one with Galileo's insight and intellectual sophistication. They are well worth reading. 95 Trajectory problems in two dimensions above a level plain. As you know, an approximation to the actual motion is obtained by assuming that the horizontal component of velocity, v z , remains constant and that the vertical component of velocity, v v , is subject to a constant acceleration of magnitude g (= 9.8 m/sec 2 ) downward. This approximate description of the motion works well provided that the effects of air resistance are unimportant, which generally speaking requires compact (dense) objects and fairly low velocities. Later we shall consider cases in which the resistive effects are important and our present- idealized picture of the motion becomes seriously inadequate. Let us now see how to answer the following questions: 1. How long is the object in flight? 2. What is the range R (the horizontal distance traveled)? 3. What is the velocity upon striking the ground? We shall choose our origin of coordinates at the starting point of the particle, and we shall take the positive coordinate directions to be upward (y) and to the righl (x). The values of the initial and final coordinates, the initial velocity components, and the acceleration components are then as shown in Fig. 3-8. Fig. 3-8 Anatysis oflhe motion o/ Fig. 3-7 in terms of separale horizontal and vertical components. p»t = t'o cos e - — *- . wmm, Horizontal Component Vertical Component < ! ox " l 'o COS W,, v„ sin o „ cos «„ «e-0 u li (minus because it is H (unknown) below the origin of the ai (always) coordinate system g (acceleration due to gravity) (a) (b) 96 Accelerated motions [A note about signs is in order here. The acceleration due to gravity is represented by a vector, g. In this book we shall use the symbol g to represent the scalar magnitude of this vector. That is, g denotes the positive quantity equal to 9.8 m/sec 2 or 32 ft/sec 2 . This is in accord with the usual convention in which the symbol A represents the positive scalar magnitude of any vector A. After a coordinate system has been chosen, and if the upward vertical direction is taken to be positive, then the y com- ponent of the vector g becomes —g (i.e., a y = —9.8 m/sec 2 ). In some situations it may be convenient to choose the downward direction as positive. In this case the y component of g must be set equal to +g. We must of course be consistent, within a given calculation, about what we mean by the positive coordinate direction, but we are completely free to take whichever choice we please; the actual content of the final answers cannot depend on this.] We return now to the trajectory problem, as depicted in Figs. 3-7 and 3-8: 1. How long is the object in flight? We know the initial velocity, the initial and final values of y, and the vertical acceleration. Therefore, if we take Eq. (3-10c), as applied to motion in the y direction, we know everything except the time of flight / : y = vo y t + \a y t 2 i.e., -h = (do sin 8 )t + \{-g)t 2 We solve this equation and take the positive root as the physically relevant one. 2. What is the range RP. Apply Eq. (3-10c) to the horizontal-component problem: x = v 0z t + h a ** 2 i.e., R = (do cos 0o)t + = (do cos 0oV In this we substitute the specific value of t obtained in cal- culation 1. 97 Trajectory problems in two dimensions 3. What is the velocity upon striking the ground? The components of v may be found from Eq. (3-10b) using the value of / obtained from calculation 1. Vz — o d 4- a z t o y = i»o» + Oyt v x = do cos Ba v v = "o sin do + (— g)' Then from these components we have Magnitude of v : o = (v z 2 + v u 2 ) l/2 o v Direction of v : tan d = — o x (In this case tan will be negative, because represents a direc- tion pointing downward below the horizontal.) Alternatively, we can calculate the magnitude of o 2 directly from Eq. (3-10b) as follows: o 2 = «Or 2 = oo 2 cos 2 0o v y 2 = v y 2 + 2a u y = o 2 sin 2 6 + 2(-*)(-A) Therefore, v 2 = o 2 + o y 2 = o 2 + 2gh The direction of v can then be calculated from the relation sin d = — v FREE FALL OF INDIVIDUAL ATOMS Atoms moving in a vacuum cast sharp shadows and give evidence of traveling in straight lines. On the other hand, it must surely be true that atoms and molecules, as samples of ordinary matter, are subject to the usual free fail under gravity at the earth's surface. The effect is not very noticeable because evaporated atoms have high average speeds— about the same as a rifle bullet — but it is measurable. How far would a beam of atoms (initially moving horizontally) fail vertically under gravity while traveling a horizontal distance L at such speeds? Figure 3-9(a) illustrates the problem. It would be just like the trajectory prob- lem of the last section, except that this time we take the horizontal distance as given and then solvc for the vertical distance y. If an origin is taken at the point O as shown, from which the atoms 98 Accelerated molions ■ / ^ (a) c — . £^— *- Fig. 3-9 (a) Trajec- o tory ofaloms in ^^^^. vacuum with an inisial horizontal velocity. The vertical displace- ment is greatly exag- gerated. {b) Para- bolic trajectory of (b) Lt >3 atoms in an atomic > beam ihat must pass Ihrough the slils A Oven and B to reach the detector D. lm- — *j« lm - D H start out horizontally with a speed v, we have Horizontal component: X = GQ x t + %a x t 2 Therefore, L = vt Vertical component: y = v 0y t + \ayt~ 2 y = + K-*)' 2 which upon substitution for / gives us y = -g L Suppose that we apply this result to a beam of atoms with a speed of about 500 m/sec. In traveling a horizontal distance of 1 m, the time of flight would be 1/500 sec (2 msec) and we should have y = — 2 \500/ 2 X 10 -5 m The deviation from a straight-line path is thus extremely small, only a few hundredths of a millimeter. Despite the small size of the effect, it has been studied with 99 Frcc fali of individual atoms precision in an experiment by Estermann et al. 1 Their arrange- ment was as shown schematically in Fig. 3-9(b). Atoms of cesium or potassium were evaporated out of an "oven" at about 450°K inside a vacuum system. Since the atoms emerge from the oven in a variety of directions, the beam was collimated by two slits A and B on the same horizontal level, as shown. The slits were about 0.02 mm wide. The beam of atoms was detected by a horizontal hot-wire detector D, also about 0.02 mm across. 2 As Fig. 3-9(b) shows, any atom that reaches the detector must have a small initial upward component velocity at A in order to negotiate the slit system. However, from a point midway be- tween A and B where the beam is horizontal, the trajectory is just like that shown in Fig. 3-9(a). The detector is moved vertically across the beam and the ion current, produced by atoms that strike the wire, is recorded. In the absence of any gravitational deflection, the intensity dis- tribution across the beam should be trapezoidal as in Fig. 3-10(a) because the central region would be bordered by "penumbra" regions that are a consequence of the two-slit system. Some results are shown in Fig. 3-10(b). The most obvious feature of these graphs is that they reveal a wide spread of speeds in the atoms of a beam. Some atoms are moving so fast that they are scarcely deflected at all; others are moving so slowly that their deflection is many times greater than the most probable deflection (which corresponds to the maximum of the intensity distribution). The complete curve must reflect a characteristic distribution of speeds of the atoms in the oven at a particular temperature. We shall not consider the detailed shape of the intensity pattern but will fix attention on the deflection of the peak. A comparison of the graphs for cesium and potassium makes it obvious that potassium atoms (atomic mass = 39) move on the 'I. Estermann, O. C. Simpson, and O. Stern, Phys. Reo., 71, 238 (1947). Similar experiments were reported at about the same time on the free fail of thermal neutrons— see L. J. Rainwater and W. W. Havens. Phys. Rev., 70, 136 (1946). 2 A neutral Cs or K. atom, striking the hot wire, becomes ionized by losing an electron. A nearby electrode at a negative potential with respect to the wire will collect these positive ions and the resultant current flow can be detected with a sensitive electrometer or galvanometer. If a thin straight wire is used, it acts as a detector of width equal to its own diameter. Some atoms (e.g., the halogens) tend to capture electrons and form negative ions; the sign of the potential of the electrode can be adjusted accordingly. 100 Accelerated motions Oven Slit/f Vertical line along which detector travels (a) *• Detector current Graph showing detector response for various positions along vertical axis 10 _ 8 8 'E 3 A i 6 .O <0 G l/l ■= 4 *- - /\ Widthof ♦f "S c 4) +-" c 4 « *-, ' 1 Width of 2 r \ o V detector 7 . >>■!-_ \_; 0.1 0.2 03 0.4 Deflection, mm (b) 0.1 0.2 0.3 0.4 Def lection, mm IPflJPI^^H^^^JIP^flU^^^^^^H F/g. 3-10 (a) Magnified detail of geometrical image formed by atoms moving in siraiglu lines through Iwo slits. The intensity is proportional t o the area ofthe source that can be "seen" by the detector in any given position. (6) Actual data on the deflection of beams ofcesium and potassium atoms. [After 1. Estermann, O. C. Simpson, and O. Stern, Phys. Rev., 71, 238 U947).] average much faster than cesium atoms (atomic mass = 133) at comparable temperatures. This is an expression of the fact that the molecules of different gases at the same temperature have equal average kinetic energies — a result that we quote without further discussion at this point. As for the actual magnitude of the thermal velocities, let us look at the peak of the cesium curve. It is displaced relative to the center of a gravitation-free beam by about 0.11 mm. Now in the arrangement shown in Fig. 101 Free fail of individual atoms 3-9(b), if A B = BC = L, then an analysis of the trajectory will show (see Problem 3-10) that atoms of speed v will be displaced downward by a distance y given approximately by y v* Hence L'- 1/2 Substituting the values L = lm and y = 1.1 X 10~ 4 m we get, for the approximatc speed, v = 300 m/sec. This very beautiful and delicate experiment was used as a test of the theoretical velocity distribution of atoms at a given temperature. ' We cite it here as a nice illustration that atoms, like baseballs or earth satellites, follow curved paths under the action of gravitational forces. Since the motion takes place in a vacuum it is, in fact, a more justifiable applicatibn of the idealized laws of free fail than are the more usual problems of objects moving through the air. You may wonder if it is possible to demonstrate the free fail of individual electrons in a similar way. This is an immensely more difficult problem, because whereas atoms, being electrically neutral, experience only the gravitational acceleration, electrons are exposed to stray electric forces that completely swamp all gravitational effects unless extraordinary precautions are taken. Nonetheless, some experiments have been attempted on this problem, although the interpretation of the results is a rather complicated affair. OTHER FEATURES OF MOTION IN FREE FALL In the idealized description of motion of a freely moving object near the earth's surface, the horizontal component of velocity always remains constant and the vertical acceleration always has the same value, g (downward). It may be interesting to point out that, under these assumptions, every trajectory associated with the same value of the constant horizontal velocity, v 0x , forms part of a single parabola — a kind of universal curve (see Fig. 3-1 1) on which one can mark in the beginning and end points of any particular trajectory, as shown. There is nothing profound 'A similar experimenl, using velocity-selected atoms, has been reported by N. B. Johnson and J. C. Zorn, Am. J. Phys., 37, 554 (1969). 102 Accelerated motions Fig. 3-11 "Universal parabola" that em- bodies all possible parabolic trajectories for a gicen downward acceleralion and a given horizontal component of velocity. The heavily marked pari ofthe curve beiween poinis C and D corresponds lo the atomic beam trajec- lory ofFig. 3-9(b). about this, but it can be useful in helping one to see any in- dividual trajectory as part of a larger scheme. For example, it makes very clear the relation between the two atomic-beam paths shown in Fig. 3-9. A closely related feature is the way in which the total velocity vector changes during the course of the motion. This is illus- trated in Fig. 3-12. Suppose that the initial velocity is repre- sented by the vector v i. Then the velocity v 2 , at a time At later, is obtained by adding to v i a vertical (downward) vector a At as shown. Similarly, every vector representing the instantaneous velocity at a subsequent stage in the motion has its end point on a vertical line drawn from the end of W\. This result embodies the fact that the horizontal component of every such vector has the same value. Yet another aspect of this same free-fall problem is illus- trated by the venerable demonstration of the hunter and the Fig, 3-12 Array of successive velocity vectors for a motion in which the accel- eralion is constant and vertically down- ward, as in free fail under gravity in the absence of air re- sistance. (This type of diagram is known as a hodograph.) aA/ 103 Other features of motion in. frce fail Fig. 3-13 Classic monkey-shooting demonsiration. The bullet and Ihe monkey undergo egual gravilational dis- placements in egual limes and are doomed to meet if ihe monkey lets go as soon as he sees ihe gun fired. monkey. The hunter aims directly at the monkey as it hangs from a limb (see Fig. 3-13). This is really a mistake, because it makes no allowance for the fact that the bullet follows a parabolic path as shown. But the monkey makes a compensating mistake. Seeing the gun aimed directly at him, he lets go of the limb as soon as he sees the flash of the gun. Thus the bullet and the monkey begin falling at the same instant (ignoring any delays due to the time of transit of the light flash and— less justifiably— the reaction time of the monkey). It then follows that, in what- ever time it takes for the bullet to travel the horizontal distance from the gun to the vertical line of the monkey's descent, both bullet and monkey receive the same contributions, \gt 2 , to their displacement as a result of the gravitational acceleration alone. Thus the bullet's trajectory crosses the line of the monkey's fail at a point that is bound to be reached at the same time by the monkey — with dirc consequences to himself. Note that this result is indcpendent of both the speed of the bullet and the value of g; it requires only that the bullet would, in the absence of gravity, go straight to the monkey's original position. Quite remarkable, on the face of it, yet easily understood in terms of the basic analysis of accelerated motion. 104 Accelerated motions Fig. 3-14 (a) Small displacement CP1P2) in a uniform circular motion. (b) Velocily veciors at the beginning and end of the shorl element ofpath. (c) Vector diagram for the evaluation of the change ofvelocity, At. UNIFORM CIRCULAR MOTION Probably the most interesting direct application of the vector definitions of velocity and acceleration is to the problem of motion in a circular path at some constant speed. In this case, if the center of the circle is chosen as an origin, the vector r always has the same length and simply changes its direction at a uniform rate. The instantaneous velocity is always at right angles to r, and its magnitude v is constant. From this we can readily calculate the acceleration. For during a short time, A/, the distance traveled is v A/, from P x to P z along a circular arc [Fig. 3-14(a)]. The angle A0 between the two corresponding directions of r is therefore given by A0 = oAt Imagine that the bisector of this angle is drawn [Fig. 3-14(b)] and consider the changes in velocity parallel and perpendicular to this bisector. Initially the velocity has a component v sin(A 0/2) away from O, and v cos(A0/2) transversely. Subsequently it has a component v sin(A0/2) toward O, and again i; cos(A0/2) trans- versely in the same direction as before. Thus the change of velocity is of magnitude 2v sin(A0/2) toward O. Figure 3-14(c) shows how this same result comes from considering a vector 105 Uniform circular motion diagram in which Av is defined as that vector which, added to v(r), gives v(/ + M). As A0 is made vanishingly small, sin(A0/2) becomes in- distinguishable from A0/2 itself (in radian measure). ' Thus we can put |Av| = 2usin(A0/2)-x;A0 But &d = v M/r, so we have |Av| = o 2 M/r Hence the magnitude of the acceleration is given by (Uniform circular motion) |a| = — (3-12) and its direction is radially inward, regardless of whether the circular path is being traced out clockwise or counterclockwise. This is called the centripetal acceleration (literally, "center- seeking") asscciated with any circular motion. The need for a dynamical means of supplying this acceleration to an object is an essential feature of any motion that is not strictly straight, be- cause any change in the direction of the path implies a component of Av perpendicular to v itself. VELOCITY AND ACCELERATION IN POLAR COORDINATES The result of the last section, and other results of more general application, are very nicely developed with the help of polar coordinatcs in the plane. The use of this type of analysis is particularly appropriate if the origin represents a center of force of some kind— e.g., the sun, acting on an orbiting planet. The starting point is to writc the position vector r as the product of the scalar distance r and the unit vector e r : r = re r (3-13) We now consider the change of r with time. This can arise from a change of its length, or from a change of its direction, or from ■Mathematically, this approximation is equivalent to (sin 6)19 -» 1 for g-,0. We shall be using this approximaiion often. For a diseussion of it, sec, for example, G. B. Thomas Jr., Calculus and Analylic Geometry, 3rd ed., Addison-Wesley, Reading, Mass., 1960, p. 172. 106 Accclcraled motions Fig. 3-15 (a) Vector change ofdisplace- ment, Ar, during a short time At in a uni- form circular motion. (b) Changes in the unit vectors t, and e« during At, showing how Ae, is parallel to et and Ae« is parallel {but opposite) to e,. a combination of both. For the present we shall limit ourselves to circular motion, in which the length of r remains constant. The change of r in a short time At is then as shown in Fig. 3- 15 (a), which is almost the same as Fig. 3-14(a). The direction of this change (Ar) is in the direction of the unit vector e$ drawn at right angles to e r as shown in Fig. 3-1 5(b). Its magnitude, as is clear from Fig. 3-15(a), is equal to r A0. Thus we can put Ar = rAdee Dividing by A/, and letting At tend to zero, we then have the result (Circular motion) dt dB v = — = r — e» dt dt (3-14a) If we designate dd/dt by the single symbol w, for angular velocity (measured in rad/sec), we have (Circular motion) v = wree = ce» (3-14b) The derivation of the above result embodies the important fact that the unit vector e r is changing with time. Although its length is by definition constant, its direction changes in accord with the direction of r itself. In fact, we can obtain the explicit expression of its rate of change as a special case of Eq. (3-14a), with r = 1 : d d6 ues (3-15a) In an exactly similar way, as Fig. 3-15(b) shows, a change of 8 implies a change of the other unit vector, ej. If the change of 6 is positive, as shown, it can be seen that the change of e$ is in the direction of — e r ; it is given by the equation 107 Velocity and accclcration in polar coordinates PROBLEMS | (e.)- -** = -we r (3-15b) This possible time dependence of the unit vectors in a polar- coordinate system is a feature that has no counterpart in rectan- gular coordinates, where the unit vectors i, j, and k are defined to have the same directions for all values of the position vector r. Once we have Eqs. (3-14a) and (3-14b) we can proceed to calculate the acceleration by taking the next time derivative. If we limit ourselves to the case of uniform circular motion, both r and to are constant, so we have (Uniform circular d , . 2 o ... a = cor — (e») = -co re r = e r motion) di r (3-16) Thus the result expressed by Eq. (3-12) falls out automatically, together with its correct direction. If we label this acceleration specifically as a radial acceleration of magnitude a r , we can put If, still restricting ourselves to motion in a circle, we remove the condition that the motion be uniform, then the acceleration vector a has a transverse component also. Starting from Eq. (3-14b), we have (Arbitrary circular do dd ,. ,_ . „ a = — e« — o — e r (3-17) motion) di dt The radial component of a is the same as we obtained for uniform circular motion (since dd/dt = v/r = «), but it is now joined by a transverse component, a«. Thus we have {_ — = — • = _ 2 r do do) d 6 ae = -r=r— =r -— di dt dfi (where co = dd/dt) (3-18) 3-1 At t = an object is released from rest at the top of a tali building. At the time /o a second object is dropped from the same point. (a) Ignoring air resistance, show that the time at which the 108 Accelerated motions objects have a vertical separation / is given by t = — + 1T glo 2 How do you interpret this result for / < £«*o 2 ? (b) The above formula implies that there is an optimum value of f o so that the separation / reaches some specified value at the earliest possible value of /. Calculate this optimum value of fo, and interpret the result. 3-2 Below are some careful measurements taken on a stroboscopic photograph of a particle undergoing accelerated motion. The distance is measured from the starting point, but the zero of time is set at the first position that could be separately identified: Time Distance (cm) (in strobe flashes) in photo 0.56 1 0.84 2 1.17 3 1.57 4 2.00 5 2.53 6 3.08 7 3.71 8 4.39 Plot a straight-line graph, based on these data, to show that they are fitted by the equation s = %a(t — to) 2 , and find /o- 3-3 A child's toy car rolling across a sloping floor is known to have a constant acceleration. Taking x = at / = 0, it is observed that the car is at x = 3 m at / = 1 sec, and at x = 4 m at / = 2 sec. (a) What are the acceleration and initial velocity of the car? (b) Plot the position of the car as a function of time up to / = 4 sec. (c) When is the car at x = 2 m? 3-4 The faculty resident of a dormitory sees an illegal water-filled balloon fail vertically past his window. Having lightning reflexes, he observes that the balloon took 0.15 sec to pass from top to bottom of his window — a distance of 2 m. Assuming that the balloon was re- leased from rest, how high above the bottom of his window was the guilty party? 3-5 The graph on the next page is an actual record of distance versus time in a straight-line motion. 109 Problems (a) Find the values of the instantaneous velocity at / = 25, 45, and 65 sec. (b) Sketch a graph of y as a function of / for the whole trip. (c) From (b) estimate very roughly the times at which the ac- celeration had its greatest positive and negative values. 3-6 In 1965 the world records for women's sprint races over different distances were as follows: 60 m 7.2 sec 100 yd 10.3 sec 100 m 11.2 sec (a) Make an accurate graph of distance in meters versus time in seconds. (b) The graph will show you that the data can be well fitted by assuming that a sprinter has a certain accelcration a for a short time t and then continues with a constant speed v. Set up the equation for distance x in terms of a, t, and /. (c) Find the numerical values of l>, a, and t that fit the de- scription given in (b). If this description is correct, what is the dis- tance in meters traveled by the sprinters before they reach their steady velocity? 3-7 Two cars are traveling, one behind the other, on a straight road. Each has a speed of 70 ft/sec (about 50 mph) and the distance be- tween them is 90 ft. The driver of the rear car decides to overtake the car ahead and does so by accelerating at 6 ft/sec 2 up to 100 ft/sec (about 70 mph) after which he continues at this speed until he is 90 ft 110 Acceleratcd motions ahead of the other car. How far does the overtaking car travel along the road between the bcginning and end of this operation? If a third car were in sight, coming in the opposite direction at 88 ft/sec (60 mph), what would be the minimum safe distance between the third car and the overtaking car at the beginning of the overtaking operation? (If you are a driver, take note of how large this distance is.) 3-8 In Paradise Lost, Book I, John Milton describes the fail of Vulcan from Heaven to earth i n the following words: . . . from M om To Noon he fell, from Noon to dewy Eve, A Summer's day ; and with the setting Sun Dropt from the Zenith like a falling Star .... (It was this nasty fail that gave Vulcan his limp, as a result of his being thrown out of Heaven by Jove.) (a) Clearly air resistance can be ignored in this trip, which was mostly through outer space. If we assume that the acceleration had the value g (9.8 m/sec 2 ) throughout, how high would Heaven be according to Milton's data ? What would have been Vulcan's velocity upon entering the top of the atmosphere? (b) (Much harder) One really should take account of the fact that the acceleration varies inversely as the square of the distance from the earth's center. Obtain revised values for the altitude of Heaven and the atmospheric entry speed. 3-9 A particle moves in a vertical plane with constant acceleration. Below are values of its x (horizontal) and y (vertical) coordinates at three successive instants of time: /, sec x, m y, m 4.914 4.054 2 X 10- 2 5.000 4.000 4 X 10~ 2 5.098 3.958 Using the basic definitions of velocity and acceleration (u x = Ax/At, etc), calculate (a) The x and v components of the average velocity vector during the time intervals to 2 X 10 -2 sec and 2 X 10~ 2 to 4 X 10 -2 sec. (b) The acceleration vector. 3-10 (a) The figure [similar to Fig. 3-9(b)] shows a parabolic atomic- beam trajectory in vacuum, passing through two narrow slits, a dis- tance L apart on the same horizontal level, and traveling an addi- tional horizontal distance L to the detector. Verify that the atoms arrive at the detector at a vertical distance v below the first slit, such 111 Problems Oven C — / *. * L * L V i ± Wall 4m that >> » gL 2 /v 2 , where u is the speed of the atoms. (You can assume y « L.) (b) A beam of rubidium atoms (atomic weight 85) passes through two slits at the same level, 1 m apart, and travels an additional distance of 2 m to a detector. The maximum intensity is recorded when the detector is 0.2 mm below the level of the other slits. What is the speed of the atoms detected under these conditions? Compare with the results for K and Cs shown in Fig. 3-10(b). What was the initial vertical component of velocity at the first slit? 3-11 (a) Galileo, in his book Two New Sciences (1638), stated that the theoretical maximum range of a projectile of given initial speed over level ground is obtained at a firing angle of 45° to the horizontal, and furthermore that the ranges for angles 45° ± 5 (where 5 can be any angle < 45°) are equal to one another. Verify these results if you have not been through such calculations previously. (b) Show that for any angle of projection (to the horizontal) the maximum height reached by a projectile is half what it would be at the same instant if gravity were absent. 3-12 A perfectly elastic ball is thrown against a house and bounces back over the head of the thrower, as shown in the figure. When it leaves the thrower's hand, the ball is 2 m above the ground and 4 m from the wali, and has i>ox - i>o» = 10 m/sec. How far behind the thrower does the ball hit the ground? (Assume that g = 10 m/sec 2 .) 3-13 A man stands on a smooth hillside that makes a constant angle a with the horizontal. He throws a pebble with an initial speed vo at an angle d above the horizontal (see the figure). (a) Show that, if air resistance can be ignored, the pebble lands at a distance j down the slope, such that s = 2v sin(0 + a) cos 8 g cos 2 <x (b) Hence show that, for given values of uo and a, the biggest value of s is obtained with = 45° — a/2 and is given by •Tmn* — t-'o (1 + sina) g COS 2 £* (Use calculus if you likc, but it is not necessary.) 112 Accelerated motions 3-14 A baseball is hit out of the stadium and is observed to pass over the stands, 400 ft from home plate, at a height of 50 ft. The ball leaves the bat at an angle of 45° to the horizontal, and is 4 ft above the ground when struck. If air resistance can be ignored (which it actually cannot), what magnitude of the ball's initial speed would be implied by these numbers? (g = 32 ft/sec 2 .) 3-15 A stopwatch has a hand of length 2.5 cm that makes one com- plete revolution in 10 sec. (a) What is the vector displacement of the tip of the hand be- tween the points marked 6 sec and 8 sec? (Take an origin of rec- tangular coordinates at the center of the watch-face, with a y axis passing upward througjt I = 0.) (b) What are the velocity and acceleration of the tip as it passes the point marked 4 sec on the dial ? 3-16 Calculate the following centripetal accelerations as fractions or multiples of g (« 10 m/sec 2 ): (a) The acceleration toward the earth's axis of a person standing on the earth at 45° latitude. (b) The acceleration of the moon toward the earth. (c) The acceleration of an electron moving around a proton at a speed of about 2 X 10 6 m/sec in an orbit of radius 0.5 A (the first orbit of the Bohr atomic model). (d) The acceleration of a point on the rim of a bicycle wheel of 26 in. diameter, traveling at 25 mph. 3-17 A particle moves in a plane; its position can be described by rectangular coordinates (x, y) or by polar coordinates (r, 0), where x = r cos and y = r sin 6, (a) Calculate a z and a y as the time derivatives of r cos and r sin 0, respectively, where both r and B are assumed to depend on /. (b) Verify that the acceleration components in polar coordinates are given by a r = a z cos 9 + a u sin 6 a» = — a x sin 6 + a„ cos 6 Substitute the values of a z and 0j, from (a) and thus obtain the general expressions for a r and a» in polar coordinates. 3-18 A particle oscillates along the x axis according to the equation x = 0.05 sin(5/ — tt/6), where x is in meters and / in sec. (a) What are its velocity and acceleration at t = 0? (b) Make a drawing to show this motion as the projection of a uniform circular motion. (c) Using (b), find how long it is, after the particle passes through the position x = 0.04 m with a negative velocity, before it passes again through the same point, this time with positive velocity. 113 Problems // seems clear to me that no one ever does mean or ever has meant by "force" rate ofchange of momentum. C. D. broad, Scientific Thought (1923) Forces and equilibrium as we said at the beginning of this book, Newton's great achieve- ment, in creating the science of mechanics, was to develop quantitative relationships between the forces acting on an object and the changes in the object's motion. More than that, he declared that the main task of mechanics was to learn about forces from observed motions. But this does not alter the fact that the idea of force exists independently of the quantitative laws of motion and comes initially from very subjective ex- periences — the muscular effort involved in applying a push or a puli. We shall begin from this point of view, and rather than plunge at once into dynamics, we shall first take a look at forces in balance. It has become rather unfashionable to do this be- cause, as you probably already know, the accepted units for the absolute measurement of force are defined in terms of the motions that unbalanced forces produce. We are bound to come to that, and in doing so we shall come to the heart of mechanics. Never- theless, the quantitative notion of force can be (and was) de- veloped in another context — the study of objects at rest, in static equilibrium. Indeed, our basic knowledge of the two most im- portant forces in mechanics — the gravitational force and the Coulomb force between electric charges — was obtained largely through laboratory observations of static equilibrium situations, using techniques that are still important and broadly applied. In discussing forces in equilibrium, we shall begin with experiences that are familiar and seemingly quite straightforward. Later, after considering some of the problems of motion, we shall recognize that tacit assumptions and unsuspected subtleties are 115 involved; we shall then be equipped to return to the problem of equilibrium with deeper insights and a broader view. But, as we said earlier, that kind of development is very good physics; we do not try to handle everything at once, but proceed by easy stages to extend the range and sophistication of our ideas. FORCES IN STATIC EQUILIBRIUI Let us consider a very simple physical system — an archer's bow. In Fig. 4-1 (a) we show the bow in its resting state; the string is straight and taut. We know that to bring the bow into the situation shown in Fig. 4-1 (b), and to hold it there, a force must be supplied. Remove that force and the bowstring snaps back toward its straight condition, perhaps launching an arrow in the process [Fig. 4-1 (c)]. What conditions have to be satisfied to hold the bow in the shape shown in Fig. 4-1 (b)? Putting aside the human, subjective aspects of the situation, one can say that the forces at the point C are in balance. But what does this mean? A force is not disembodied; it is applied to something. Fig. 4-1 (a) Schematic diagram of an archer's bow in resting stale. (b) Bowstring drawn back at midpoint, regu/ring applied force along the bisector ofthe angle between the two segments of string. (r) Arrow in the pro- cess ofbeing launched. (d) Forces applied at the center of the bowstring in situation (6). 116 Forces and equilibrium Fig. 4-2 (a) Balanc- ing of equal forces in simple egual-arm ar- rangement. (b) Bal- aiicing ofunequal forces in accordance with the law ofthe lever. S l> O (a) o S S ^> S fe f l (b) One cannot imagine a force in the absence of a physical object on which it is exerted. In this case the object in question is some small part of the bowstring in the immediate neighborhood of point C. This little piece of string is exposed to pulls along directions CA and CB from the adjoining portions of the string, and to a third force, supplied by the archer, along the bisector of angle ACB. We can draw a separate diagram, as in Fig. 4-1 (d), showing the piece of string at C in equilibrium under the action of three forces — a force F from the archer, and two forces, which by symmetry are of equal magnitude T, from the string. These latter combine to give a force, equal and opposite to F, along the line of symmetry. But how do we know that the equi- librium requires that the bowstring supply a net force equal and opposite to F? You might say that this is obvious, but we can back up this intuition with a real experiment. It will involve one assumption: that identical objects, equally deformed, supply pulls or pushes of cqual size. For example, a small loop of string can remain at rest if pullcd in opposite directions by two identical coiled springs extended by equal amounts. And this balancing of equal and opposite forces is the most elementary of all equi- librium situations. Problems such as that of the archer's bow will not be new to you. But let us point out that the analysis of them involves our ability to assign numerical magnitudes to individual forces and to compare one force with another. How can we do that? Archimedes showed the way, when he discovered the law of the lever over 2000 years ago. Equal forces balance when applied at equal distances on either side of the pivot O [Fig. 4-2(a)]. Unequal forces balance [Fig. 4-2(b)] if the condition F\l\ = F 2 h is satisfied. Considerations of symmetry alone would suggest the first result, but the second has to be based on experi- ment. ' To establish it we need to have a way of obtaining ■Archimedes believed that he could obtain the result by pure logic, but this is not so. See Problem 4-5. 117 Forces in static equilibrium Fig. 4-3 Basic ar- rangement of a "steel- yard" for weighing an unknown (X) with the help of a Standard weight (S) that can be moved along the horizontal arm. multiples of a given force. One method would be to construct a number of identical coiled springs. We could first verify that, when stretched by equal amounts, they would balance one another individually in an equal-arm arrangement. Then we could attach several, all at the same distance on one side of the pivot, and balance them with a single spring, or some different multiple, at the appropriate distance on the other side, as shown in Fig. 4-2(b). (Note that such a procedure entails no assump- tions about the way in which the force varies with extension for an individual spring.) Once the law of the lever has been justified in this way, we can use it to measure an unknown force in terms of an arbitrary Standard. The technique has been used since Archimedes' time, at the very least, for the purpose of weighing objects (see Fig. 4-3), and its basic principle, which allows us to balance a small force against a large one, using suitably chosen lever arms, is exploited in many familiar mechanical devices. UN1TS OF FORCE For the purposcs of this introductory study of forces in equi- librium, we do not strictly necd to worry about the absolute magnitudes of forces. The introduction of an arbitrary unit of force, and the specification of other forces as multiples of this unit, would really suffice. However, it will often be convenient to express forces in terms of their customary measures, so before going further we shall state what these measures are. (You have probably met them all before, in any case.) The unit of force that we shall most frequently employ, here and throughout the book, is the newton (N). This is a force of 118 Forces and equilibrium such a size that it can give an acceleration of 1 m/sec 2 to a mass of 1 kg ; it represents the basic unit of force in the MKS system. The detailed basis of this definition will be discussed in Chapter 6; all that we need for the moment is the recognition that this does uniquely define a force of a certain size. The magnitude of this unit is such that the gravitational puli exerted on a mass of 1 kg near the earth's surface is about 9.8 N. A force of about 1 N is represented by the earth's puli on a medium-sized apple — a most appropriate result in view of the old tradition (which may well be true) that the fail of an apple provided the starting point of Newton's profound thoughts about gravitation. In the CGS (centimeter-gram-second) system, the unit of force is the dyne (dyn), defined as the force that can give an acceleration of 1 cm/sec 2 to a mass of 1 g. Since 1 cm = 10~ 2 m, and 1 g = 10 -3 kg, we have the relationship 1 dyn = 10- 5 N (In saying this we take the relation F = ma as already estab- lished, so that changing m and a by given factors implies changing F by the product of those factors.) The dyne is thus an ex- ceedingly small force, about equal to the earth's gravitational puli on a mass of 1 mg — as represented, for example, by roughly a £-in.-square piece of this page. The only other force that we shall have occasion to mention is the pound. Unlike the newton and the dyne, the pound is (or at least was, originally) based directly on the measure of the earth's gravitational puli on a Standard object. As soon as one recognizes that the gravitational puli on any given object changes from one place to another, this definition loses its exactness and has to be adjusted. However, setting aside such difnculties of detail, we can still say that the pound is a force equal in mag- nitude to about 4.5 N. Every time we buy something that is weighed out on a spring scale, we are accepting the use of gravita- tional units of force such as the pound. We shall considcr the practical implications of that in more detail later (Chapter 8). EQUILIBRIUM CONDITIONS; FORCES AS VECTORS The static equilibrium of a given object entails two distinct conditions: 1 . The object shall not be subject to any net force tending to 119. Ecjuilibrium conditions; forccs as veclors Fig. 4-4 (a) An ob- ject in ekuilibrium uuder the action of three nonparailel forces. (6) Two equally acceptable vector diagrams show- ing the eguilibrium condition 2F = 0. (a) (b) move it bodily; it is in what we call translational equilibrium. 2. The object shall not be subject to any net infiuence tending to twist or rotate it; it is in what we call rotational equilibrium. The first condition involves, in general, the combination of forces acting in different directions, as in our initial example of the bowstring. It has been known since long before Newton's time that forces are vectors. This says much more than that they have characteristic directions; it says that they combine with one another in the same way as the prototype vector quantity, posi- tional displacement. Imagine, for example, a ring that is a loosc fit over a vertical peg. Suppose that it has three strings attached to it and that these strings are pulled by forces of relative mag- nitudes 3, 4, and 5, as defined by corresponding numbers of identical springs equally extended [see Fig. 4-4(a)]. Then ex- periment will show that the ring remains in equilibrium, even when the peg is removed, if the directions of the forces cor- respond exactly to those of a 3-4-5 triangle. This means that the forces, represented as vectors with lengths proportional to their magnitudes, form a closed triangle, which can be drawn in two different ways, as shown in Fig. 4-4(b). In formal terms, we say that the three force vectors add up to zero: Fi +F2 + F3 = It is a basic property of vectors that the order of addition is immaterial (Chapter 2). Thus, if we have a large number of forces applied to the same object, it is possible to represent their addition in many different ways; Fig. 4-5 gives an example. The one essential feature is that, in every case, the force vectors form 120 Forces and equilibrium Fig. 4-5 (a) Seueral forces acting at the same poin:, (b) The force veciorsform a closed polygon, show- ing equilibrium. (c) Equioalent veclor diagram lo (b). a closed polygon (i.e., they add up to zero) if equilibrium exists. Since forces may be applied in any direction in three-dimensional space, the force polygon is not necessarily confined to a plane, and the single statement that the force vectors add up to zero will in general be analyzable into three separate statements per- taining to three independently chosen directions — usually, al- though not necessarily, the mutually orfhogonal axes of a rec- tangular coordinate system. Geometrically, one can think of this as the projection of the closed vector polygon onto different planes; regardless of the distortions of shape, the projected polygon remains a closed figure. When written out in algebraic terms, the projection involves a statement of the analysis of an individual force vector into components, or resolved parts, along the chosen directions. Thus the first condition of equilibrium — equilibrium with respect to bodily translation — can be written as follows: Vector statement: F = Fi +F2 + F3 + ••■ = Component statement: F z = F U + F 2l + F 3l + ■ ■ ■ = F u = Fi„ + F 2u + Fa, + ■ ■ ■ = F z = Fu + F 2l + F 3l + • • • = (4-la) (4-1 b) It is worth noticing the relationship between the process of adding force vectors and the process of resolving an individual force vector into its components. I n Fig. 4-6(a) we show a force F, in the xy plane, resolved into the usual orthogonal components: F = Fj + F„j 121 Equiiibrium conditions; forces as vectors Fig. 4-6 (a) Re sol u t ion ofaforce inlo orthogonal com- ponents. (b) The components of F, added vectorially to theforce —F, giue zero. (c) Resoluiion of F inlo nonorthogonal components. We represent F by a full line and its components by dashed lines, to emphasize that the components are a replacement — a sub- stitute— for F itself. In Fig. 4-6(b) we show a elosed veetor foree triangle, in which the veetors F x \ and F„\, added to the foree veetor —F, represent an cquilibrium situation: FA + F v i + (-F) = We can then recognize that any other pair of veetors that give zero when added to — F are a complete equivalent to F itself. Thus we have at our disposal an infinite variety of ways of re- solving a given foree into components. For example, the two forees F m > and F v > in Fig. 4-6(c) could represent the way of analyzing F into components in an oblique coordinate system; the faet that F,< may happen to be larger than F itself does not invalidate this way of resolving the original foree. In a similar vein, if any set of forees combine to give zero, as in Fig. 4-7(a), we can regard one of them (which we have Fig. 4-7 (a) The combination of forees Fi, F 2 , and F 3 is bal- anced by the equil- ibrantF B . (b) The same combination of forees Fi, F2, and F3 has the resultant F K . 122 Forees and equilibrium labeled here as ¥ E ) as the equilibrant (i.e., the balancer) of all the others. A force equal and opposite to F E is then a force completely equivalent to the vector sum of these other forces: it is their resuJtant, shown as ¥ R in Fig. 4-7(b). A familiarity with these simple relationships can be a considerable help in problems of combining or resolving forces and can strengthen one*s understanding of equilibrium situations in general. ACTION AND REACTION IN THE CONTACT OF OBJECTS We have already used a kind of principle of uniformity to pre- scribe a way of reproducing a force of a given magnitude or making a known multiple of the force. We assume that identical springs, stretched or compressed by equal amounts, define equal forces. This assumption leads to consistent conclusions in the analysis of equilibrium situations. Picture now a simple ex- periment that any two people can do with a pair of identical springs. They stand facing each other, and one of them, A, agrees to be the active agent. His goal is to push (via the spring that he holds) on the end of the spring held by B, the passive partner, who tries to avoid pushing back. The size of the push that each person exerts is measured by the amount by which his spring is compressed. You know the result of the experiment, of course. Regard- less of the maneuvers made by A and B, the compressions of the two springs are always the same ; only the amount of both com- pressions together can be varied. This displays, in a particularly direct manner, the phenomenon that is expressed by the familiar statement: Action and reaction are equal and opposite. In its general form this says that, regardless of the detailed form of the contact or of the relative hardness or softness of the two objects involved, the magnitudes of the forces that each object exerts on the other are always exactly equal. Note particularly that, from the very way they are defined, these two forces cannot both act on the same object. This may seem a trivial and obvious remark, but many calculations in elementary mechanics have come to grief through a failure to recognize it. The production of a force of reaction in response to an applied force always involves deformation to some extent. You push on a wali, for example, and that is a conscious muscular act; but how does the wali know to push back? The answer is that it 123 Action and reaction in the contact of objects yields, however imperceptibly, and it is as a result of such elastic deformations that a contact force exerted by the wali comes into existence. No matter how rigid a surface may sefem, it always gives a little under a push or a puli and cannot supply a contrary force until it has done so. There is no basic difference between what happens when we sit down in a comfortably upholstered chair and what happens when we sit down on a concrete ftoor. But in the one case the springs are soft and yield visibly by distances of an inch or more, whereas in the other case the springs are essentially the individual atoms in a tightly packed solid structure, and a deformation by a small fraction of an atomic diameter is enough to produce the support required. ROTATIONAL EQUILIBRIUM; TORQUE We shall now consider the second condition for static equilib- rium of an object, assuming that the first condition, £F = 0, has been satisfied. Whether the object is, in fact, in equilibrium will now depend on whether or not the forces are applied in such a way as to produce a resultant twist. Figure 4-8 illustrates the problem with the simplest possible example. An object is acted on by two equal and opposite forces. If, as in Fig. 4-8(a), the forces are along the line joining the points A and B at which the forces are applied, the object is truly in equilibrium; it has no tendency to rotate. In any other circumstances [e.g., as shown in Fig. 4-8(b)] the object is bound to twist. If the directions of the forces remain unchanged as the object turns, equilibrium orienta- tion is finally reached, as shown in Fig. 4-8(c). How do we con- Fig. 4-8 (a) Rota- tional equilibrium with equal, opposite forces applied at different points of an extended object. (b) Equal, opposite forces ap- plied in a way that does not give rota- tional eouilibrium. (c) Iffree to rotate, the object mouesfrom orientation (b) to an equilibrium orienta- tion. 124 Forces and equilib.rium Fig. 4-9 Alt the forces acling on a pivoted bar (pf negligible weight) in rolalional equilib- rium with one force applied on each side ofthe pivot. F,+F, struct out of such familiar knowledge a quantitative criterion for rotational equilibrium? The law of the lever provides the clue. Look again at the situations shown in Fig. 4-2. In particular, consider situation (b). The balancing of the forces F t and F 2 with respect to the pivot at O reauires the condition FJi = F 2 / 2 . The product of the force and its lever arm describes its "leverage," or twisting ability; the technical term for this is torque. The torques of F t and F 2 with respect to O are equal in magnitude but opposite in direction — that due to F 2 is clockwise and that due to F\ is counterclockwise. Let us call one of them positive and the other one negative; then the condition of balance can be expressed in another way: The total toraue is equal to zero. Although the situation as described above is extremely simple, there is more to it than meets the eye, because a further force is exerted on the bar at the position of the pivot ; it must be of magnitude F, + F 2 if the first condition of equilibrium is to be satisfied. Thus the complete array of forces is as shown in Fig. 4-9. Now, to be sure, this third force exerts no torque about the pivot point O itself. However, what if we choose to consider the torques about, let us say, the left-hand end of the bar, some distance d to the left of the point of application of F x ? Then clearly, with respect to this new origin, the force at O is supplying a counterclockwise torque — but it turns out that this is exactly balanced by the sum of the torques due to Fj and F 2 (both clockwise, notice, with respect to the new, hypothetical pivot), provided that the condition F\l\ = F 2 / 2 is satisfied. If this result is new to you, take a momcnt to convince yourself of its correctness. What it says is that, if the vector sum of the forces on an object is zero, and if the sum of the torques about any one point is zero, then the sum of the torques about any other point is also zero. So far we have limited ourselves to the balancing of torques of parallel forces. Now let us make things more general. Suppose that a force F is applied at a point P, somewhere on or in an 125 Rotational equilibrium; torquc K F sin + \Sf COS ip l ' (J* ( = r sin^) /v (a) (b) Rgi 4-/0 (a) Force F applied at a vector distance r front a pivot point. (6) Resolution of F into components along and perpendicular lo r. (c) Evaluation of lorgue of F by finding its effeclive lever arm, I. object [Fig. 4-10(a)]. Consider the torque produced by F about some other point O, which might be the position of a real pivot, or just an arbitrary point. Let the vector distance from O to Z' be r. The first thing to notice is that r and F between them define a plane, which we have chosen to be the plane of the diagram. Experience, so familiar that it has become second nature, tells us that if O were indeed a real pivot point, the effect of F would be to produce rotation about an axis perpendicular to the plane in which r and F lie. It therefore makes excellent sense to associate this direction with the torque itself, regarded as a vector of some sort. Now, what about the magnitude of the torque? We can calculate this in two ways. The first, indicated in Fig. 4-10(b), is to resolve the force into components along and perpendicular to r. If the angle between r and F is <p, these components are Fcos<p and F sin v», respectively. The radial component represents a force directed straight through O and hence contributes nothing to the torque. The transverse com- ponent, perpendicular to r, gives a torque of magnitude rFsin <?. Another way of seeing this result is, as suggested in Fig. 4-10(c), to recognize that the effective lever arm of the total force F can be formed by drawing the perpendicular ON from O to the line along which F acts. Then the torque due to F is just the same as if it were actually applied at the point W, at right angles to a lever arm of length / equal to r sin <p. ' We shall introduce the single symbol M for the magnitude •Leonardo da Vinci envisaged the effeclive lever arm of a force in this way and called it "the spiritual distance of the force." 126 Forces and equilibrium Fig. 4-11 {a) Cross product. C, oftwo arbilrary vectors, A and B. (b) Torque vector, M, as the cross product oft and F. (c) Silualion resulling in a torque vecior opposile in direclion to thal in (b). of the torque. Then we have M = rF siri <p (4- 2a ) This equation does not contain the necessary information about the direclion of the torque, but a compact statement in vector algebra, invented specifically for such purposes, is at hand. This is the so-called cross product or vector product of two vectors. Given two vectors A and B, the cross product C is defined to be a vector perpendicular to the plane of A and B and of magnitude given by C = AB sin 6 where is the Iesser of the two angles between A and B (the other being 2v — d). There are, of course, two opposite vector directions normal to the plane of A and B. To establish a unique convention we proceed as follows. Imagine rotating the vector A through the (smaller) angle until it lies along the direction of B [see Fig. 4-1 l(a)]. This establishes a sense of rotation. If the fingers of the right hand are curled around in the sense of rota- tion, keeping the thumb extended (do it!), the direction of the cross product is along the direction of the pointing thumb. The shorthand mathematical statement, which is understood to em- body all these properties, is then written as C = A X B 127 Rolational equilibrium; torque Note carefuily that the order of the factors is crucial; reversing the order reverses the sign : B X A = -(A X B) Using this vector notation, the torque as a vector quantity is completely specified by the following equation: M = r X F (4-2b) Figure 4- 11 (b) and (c) illustrates this for two different values of <p; in each case a right-handed rotation about the direction in which M points represents, as you can verify, the direction in which F would cause rotation to occur. Then, finally, we can write down the vector sum of all the torques acting on an object, and the second condition of cquilibrium — equilibrium with respect to rotation— can be written as follows: YM = ri X Fi + r 2 X F 2 + r 3 X F 3 + • • • = (4-3) If the object on which a set of forces acts can be regarded as an ideal particle (i.e., a point object), then the condition of rotational equilibrium becomes superfiuous. Since all the forces are applied at the same point, they cannot exert a net torque about this point; and if the condition £F = is also satisfied, they cannot exert a net torque about any other point either. If one wants to put this in more formal terms, one can say that the same value of r applies to every term in Eq. (4-3), so that the condition £M = reduces to the condition r X (EF) = and so embodies the condition £F = o for translational equi- librium; the equation for rotational equilibrium adds no new information. FORCES WITHOUT CONTACT; WEIGHT Figure 4-12 depicts a pair of situations that look so simple, and embody results that are so familiar, that you may never have paused to wonder about the relation between them. In Fig. 4-1 2(a) an object is shown attached to two spring balances that puli on it horizontally with equal and opposite forces— a clear case of static equilibrium. In Fig. 4-1 2(b) the same object hangs 128 Forces and equilibrium -~n=\ — # — E>^ Fig. 4-12 (o) Objecl in equi- librium under equal and opposite horizontal forces (from spring balances). (g) {b) Objecl in eauilibrium under the action ofa single uerlical contaclforce (from a spring balance) and an incisible in- fluence (gracity). (c) The reading on the spring balance in (b) defines the measured weight (W) of the objecl. Byinference the gravitational force F„ is (b) (C) equal and opposite to W. from a single spring balance that pulls vertically upward on it. Experience tells us that this is also a case of static equilibriutn, yet as far as visible connections are concerned, it is quite clearly an unbalanced situation. How do we resolve the dilemma? This may seem like a trite or even trivial question. In fact, it is a very profound one. At this time we shall examine it only from the standpoint of our knowledge of static equilibrium. Later we shall see that it has much wider ramifications. Try to imagine coming to situation (b) of Fig. 4-12 for the first time, with a background of experience in other kinds of static equilibrium. You have learned from this previous ex- perience that equilibrium always corresponds to having the vector sum of the applied forces equal to zero. You are used to having tangible and visible evidence of these forces being applied, via strings, springs, and so on. Now you see an equilibrium situation in which only one force is i n f uli view, so to speak. But your confidence in the general validity of the equilibrium con- ditions is so strong that you say: "Even though there is no other contact, there must be another force acting on the object, equal and opposite to the force supplied by the spring." And so you postulate the force of gravity, pulling downward on the object. But your only measure of it is the reading on the spring balance that counteracts this gravitational force. It is in terms such as these that we have come to postulate and accept the existence of a downward gravitational force exerted on every object at or near the earth's surface. In order to hold an object at rest relative to the earth, we must apply a certain upward force on it. The magnitude of this equilibrating force, as measured for example on a spring balance, is what we shall call the weight of an object. This is an important definition; 129 Forces without contact; weight we shall restate it : The weight of cm object will be defined as the magnitude, W, of the upward force that must be applied to the object to hold it at rest relative to the earth. This is an example of what is called an operalional dcfinition; we describe the actual process that is to be followed to get a practical measurement of the quantity in question. Notice especially that we are not defining the weight of an object as the gravitational force on it. If the connection with the supporting spring is broken, so that the object begins to fail freely, then by our definition the object is now weightless, although there is no suggestion that the gravi- tational force on the object has been changed in any way in the process of breaking the connection. (We shall diseuss weight- lessness further in Chapter 8.) If we return to the situation in which the object is held stationary relative to the earth by the puli of the spring, then our picture of the forees aeting on the object is as shown in Fig. 4-l2(c). On the assumption that this is indeed a static equilibrium situation, then we can say that the spring force W and the gravitational force F„ are equal and opposite, so that the magnitude of W does provide, under these cireumstances, a measure of the magnitude of F„. But keep it firmly in mind that, in our usc of the terms, weight and gravitational force are not synonymous. By maintaining this distinetion we shall be much better equipped to handle dynamical problems involving gravity later on. PULLEYS AND STR1NGS The use of pulleys and strings to transmit forees has more physics in it than one might think and contains some nice applications of the principles of static equilibrium. Consider a string passing over a cireular pulley of radius R [Fig. 4-1 3(a)]. Let the string be pulled by forees F i and F 2 as shown. If this is to be a situation of static equilibrium, the pulley must be in both translational and rotational cquilibrium. Let us consider the rotational equi- librium first. If the pivot is effectivcly frictionless, the only torques on the pulley are supplied by the forees Fi and F 2 . Since both of these are applied tangcntially, they have equal lever arms of length R (assuming that the pulley is pivoted exactly at its center). Therefore, to give zero net torque, the magnitudes of Fj and F 2 must both be equal to the same value, T, which we can thus call the tension in the string — the strength 130 Forees and eqiiilibrium Fig. 4-13 (o) Ten- sions in the segments of a st ring where it meels a slationary circular pulley must be in rotational equi- librium. (b) Set of three forces involced in the static eguilib- rium of a pulley. (c) Pulley as a deuicefor applying a force of gicen magnitude in any desired direction. (a) (b) (O of the puli that would be found to be exerted by the string at any point along its length. The pulley thus enables us to change the direction of an applied force without changing its magnitude. However, to satisfy the condition of translational equilibrium, the pivot must be able to supply a force F 3 equal and opposite to the vector sum of F, and F 2 . This force must therefore lie along the bisector of the angle between the two straight seg- ments of the string, and its magnitude must be equal to 2Tcos(6/2) [see Fig. 4-13(b)]. If the tension at one end of a string passing over a pulley is supplied by a suspended object, then a force F of magnitude equal to the weight, W, of the object is needed to maintain equi- librium [Fig. 4-13(c)]. This means that a pulley-string system can be used to supply a force of magnitude W in any desired direction. Figure 4-14 illustrates the typical kind of arrangement that exploits these features in a simple experiment to study the equilibrium of three concurrent forces. To the extent that the pulleys can be treated as ideal, the magnitudes of the forces Fi, Fig. 4-14 (a) Simple static equilibrium ar- rangement involving three nonparallel forces (string ten- sions) applied at the pomt P. (b) Vector diagram showing the equilibrium condition. (a) (b) 131 Pulleys and strings PROBLEMS K (a) H l- H-. -I' (b) F 2) and F 3 are equal to the respective weights W\, W 2 , and W 3 of the suspended objects. 4-1 The ends of a rope are held by two men who puli on it with equal and opposite forces of magnitude F. Construct a clear argument to show why the tension in the rope is F, not 2F. 4-2 It is a well-known fact that the total gravitational force on an object may be represented as a single force acting through a uniquely defined point— the "center of gravity"— regardless of the orientation of the object. (a) For a uniform bar, the center of gravity (CG) coincides with the geometrical center. Use this fact to show that the total gravita- tional torque about the point P [see part (a) of the figure] may be considered as arising from a single force W at the bar's center, or from two individual forces of magnitudes Wx/L and W(L — x)/L acting at the midpoints of the two scgments defined by P. (b) If a bar or rod has a weight W, and a small weight w is hung at one end [see part (b) of the figure], use the simpler of the above two methods to show that the system balances on a fulcrum placed atPif x = LW/UW + »v). 4-3 Diagram (a) represents a rectangular board, of negligible weight, with individual concentrated weights mounted at its corners. 3 1 H'. H», > iy O w, 'W, (a) (b) (a) To find the position of the CG of this system, one can proceed as follows: Choose an origin at the corner O, and introduce x and y axes as shown. Imagine the board to be pivoted about a horizontal axis along y, and calculate the distance x from this axis at which an upward force W (= wi +K-2-f-H'3 + w 4) will keep the 132 Forces and equilibrium system in rotational equilibrium. Next imagine the board to be pivoted about a horizontal axis along x, and calculate the correspond- ing distance y. Then the center of gravity, C, is at the point (x. y). (b) An experimental method of locating the CG is to hang the board from two corners in succession (or any other two points, for that matter) and mark the direction of a plumbline across the board in each case. To verify that this is consistent with (a), imagine the board to be suspended from O in a vertical plane [diagram (b)] and show by direct consideration of the balancing of torques due to W2, h% and m>4 that the board hangs in such a way that the vertical line from O passes through C (so that tan d = y /2). 4-4 You have just finished a 20-page letter to your giri friend (or boyfriend) and you want to mail it at once, so that it will be collected first thing in the morning. You have a supply of stamps, but un- fortunately it is 2 a.m., the local post office is closed, and you haven't a letter scale of your own. However, you do have a 12-in. ruler and a nickel, and you happen to have learned somewhere that the density of nickel is about 9 g/cm 3 . The ruler itsclf balances at its midpoint. When the nickel is placed on the ruler at the 1-in. mark, the balance point is at the 5-in. mark. When the letter is placed on the ruler, centered at the 2-in. mark, the balance point is at S^in. The postal rate is 20 cents per half-ounce (international airmail). How much postage should you put on? (This problem is drawn from real life. You might like to do a similar experiment for yourself, perhaps using a differcnt coin as your Standard of weight.) 4-5 (a) As mentioned in the text (p. 1 17), Archimedes gave what he believed to be a theoretical proof of the law of the lever. Starting from the necessity that equal forces, F, at equal distances, /, from a fulcrum must balance (by symmetry), he argued that one of these forces could, again by symmetry, be replaced by a force F/2 at the fulcrum and another force F/2 at 21. Show that this argument depends on the truth of what it is purporting to prove. (b) A less vulnerable argument is based on the experimental 133 Problems (b) £- r knowledge that forces combine as vectors. Suppose that parallel forces F\ and F2 are applied to a bar as shown. Imagine that equal and opposite forces of magnitude / are introduccd as shown. This gives us two resultant force vectors that intersect at a point that defines the line of action of their resultant (of magnitude Fi + F2). Show that this resultant intersects the bar at the pivot point for which Fili = F2/2. 4-6 Find the tensions in the ropes in the two configurations shown. The weight W is in static equilibrium in each case. 4-7 (a) A elothesline is tied between two poles, 10 m apart, in such a way that the sag is negligible. When a wet shirt with a mass of 0.5 kg is hung at the middle of the line, the midpoint is pulled down by 8 cm. What is the tension in the elothesline? (b) (With acknowledgements to F. W. Sears.) A car is stranded in a diteh, but the driver has a length of rope. The driver knows that he is not strong enough to puli the car out directly. Instead, he ties the rope tightly between the car and a tree that happens to be 50 ft away ; he then pushes transversely on the rope at its midpoint. If the midpoint of the rope is displaced transversely by 3 ft when the man pushes with a, force of 500 N (= 50 kg ), what force does this exert on the car? If this were suflicient to begin to move the car, and the man pushed the midpoint of the rope another 2 ft, how far would the car be shifted, assuming that the rope does not streteh any further? Does this seem like a practical method for dealing with the situation? 4-8 Prove that if three forces act on an object in equilibrium, they must be coplanar and their lines of action must meet at one point (unless all three forces are parallel). 4-9 Painters sometimes work on a plank supported at its ends by long ropes that pass over fixed pulleys, as shown in the figure. One end of each rope is attached to the plank. On the other side of the pulley the rope is loopcd around a hook on the plank, thus holding the plank at any desired height. A painter weighing 175 lb works on such a plank, of weight 50 lb. (a) Keeping in mind that he must be able to move from side to side, what is the maximum tension in the ropes? (b) Suppose that he uses a rope that supports no more than 150 lb. One day he finds a firm nail on the wali and loops the rope around this instead of around the hook on the plank. But as soon as he lets go of the rope, it breaks and he falls to the ground. Why? 4-10 An inn sign weighing 100 lb is hung as shown in the figure. The supporting arm, freely pivoted at the wali, weighs 50 lb, and the system is held by a guywire that should not be subjected to a tension of more than 250 lb. 134 Forces and equilibrium * Slr Isaac Newton £ ift 2ft 2ft M i i w (a) What is the minimum safe distance of the point B above poin t A ? (b) What is the magnitude and direction of the force exerted on the supporting arm at A under these conditions? 4-11 A man begins to climb up a 12-ft ladder (see the figure). The man weighs 180 lb, the ladder 20 lb. The wali against which the ladder rests is very smooth, which means that the tangential (vertical) com- ponent of force at the contact between ladder and wali is negligible. The foot of the ladder is placed 6 ft from the wali. The ladder, with the man's weight on it, will slip if the tangential (horizontal) force at the contact between ladder and ground exceeds 80 lb. How far up the ladder can the man safely climb? 4-12 You want to hang a picture at a certain place on a wali, but the only available nails are at points 1 ft to the left and 2 ft to the right of the edges of the picture (see the figure). You attach strings of the appropriate lengths from these nails to the top corners of the picture, as shown, but the picture will not hang straight unless you add a balancing weight of some kind. (a) If the picture with its frame weighs 10 lb, what is the least possible balancing weight, and where would you put it ? (Hint: Find the point of intersection of the two tension forces in the strings.) (b) In the absence of the balancing weight, how would the picture hang? (If you want to go beyond a qualitative discussion, be prepared for some rather messy trigonometry.) 4-13 A yo-yo rests on a table (see the figure) and the free end of its string is gently pulled at an angle 6 to the horizontal as shown. (a) What is the critical value of 6 such that the yo-yo remains stationary, even though it is free to roli? (This problem may be solved geometrically if you consider the torques about P, the point of contact with the table.) 135 Problems (b) What happens for greater or lesser values of 0? (If you have a yo-yo, test your conclusions experimentally.) 4-14 A simple and widely used chain hoist is based on what is callcd a diffcrential pulley. In this arrangement two pulleys of slightly different diameter are rigidly connected with a common, fixed axis of rotation. An endless chain passes over these pulleys and around a free pulley from which the load W is suspended (see the figure). If the components of the ditferential pulley have radii a and 0.9a, respec- tively, what downward puli applied to one side of the freely hanging part of the chain will (ignoring friction) suffice to prevent the load from descending if (a) the weight of the chain itself can be neglected? (b) (more realistic) one takes account of the fact that the freely hanging portion of the chain (PQR) has a total weight w? 4-15 A force F with components F z — 3N, F y = AN, and F z = is applied at the point x = 0, y = 5 m, and z = 4 m. Find the mag- nitude and direction of the torque, M, of F about the origin. (Express the direction in terms of the direction cosines — i.e., the cosines of the angles that M makes with the axes.) 4-16 Analyze in qualitative but careful terms how the act of pushing vertically downward on the pedal of a bicycle results in the production of a horizontal force that can accelerate the bicycle forward. (Clearly the contact of the rear wheel with the ground plays an essential role in this situation.) 136 Forees and equilibrium And thus Nature will be very conformable to herselfand very simple, performing all the great Motions of the heavenly Bodies by the attraction of gravity . . . and almost all the small ones of their Particles by some other attracting and repelling Powers .... newton, Opticks (1730) The various forces of nature THE BASIC TYPES OF FORCES 1 all forces arise from the interactions between difFerent objects. Once upon a time it must have seemed that these interactions were bewilderingly diverse, and one of the most remarkable features in the development of modern science has been the growing realization that only a very few basically distinct kinds of interaction are at work. The following are the only forces that we know of at present: 1 . Gravitational forces, which arise between objects because of their masses. 2. Electromagnetic forces, due to electric charges at rest or in motion. 3. Nuclear forces, which dominate the interaction between subatomic particles if they are separated by distances less than about 10 -15 m. It may be that evcn this degree of categorization will prove >An excellent background to this topic is the PSSC film, "Forces," by J. R. Zacharias, produced by Education Development Center, Inc., Newton, Mass., 1959. The title of this chapter is borrowed from that of a set of popular lec- tures delivered in London by the great scientist, Michael Faraday, just over 100 years ago and available in paperback (Viking, New York, 1960). They make easy and rather delightful reading. 139 to be unnecessarily great; the theoretical physicist's dream is to find a unifying idea that would allow us to recognize all these forces as aspects of one and the same thing. Albert Einstein spent most of his later years on this problem but to no avail, and at the present time the assumption of several different kinds of forces seems meaningful as well as convenient. In the following sections we shall briefly consider these three primary types of forces, with examples of physical systems in which they are significant. It will be useful, and from the standpoint of classical mechanics very important, to add to our classification what we shall call "contact forces" — the forces manifested in the mechanical contact of ordinary objects. Al- though these forces are merely the gross, large-scale manifesta- tion of the basic electromagnetic forces between large numbers of atoms, they serve so well to describe most of the familiar interactions in mechanical phenomena that they merit a category of their own. GRAVITATIONAL FORCES Ali our experience suggests that a gravitational interaction be- tween material objects is a universal phenomenon. It is always an attractive interaction. The gravitational forces exerted by the earth on different objects near its surface can be compared, by using a spring balance for example, and these gravitational attractions are proportional, in every case, to the property of the attracted object that we call its mass. (The content and the im- plications of this familiar and seemingly simple statement will be a matter for detailed discussion in later chapters.) The general law of gravitational interaction arrived at by Newton states that the force F with which any particle attracts any other is proportional to the product of the masses of the particles, inversely proportional to the square of thcir separation, and directed along the line separating the two particles. It is found experimentally that this force has no measurable de- pendence on the velocity of the particle on which it acts. ' The magnitudo of the force F 1 3 that a particle of mass m 1 exerts on a ■Departures from ihis velocity indcpendence are analyzcd in the general theory of relativity. They are discernible only if the effect of quantities of the order of o 2 /c 2 (where c is the speed of light) can be detected in the gravi- tational interaction. 140 The various forces of natura particle of mass m 2 can be written as F l2 = G^f (5-1) where r l2 is the distance from the center of mi to the center of m 2 and G is a constant of proportionality called the universal gravitational constant. This law of force holds for point masses and for uniform spheres of finite size. The value of the constant G is 0.667 X 10- lo m 3 /kg-sec 2 . Equation (5-1), and the preceding verbal expression of it, is the first example in this book of the quantitative expression of a physical law. It is worth spending a few words to discuss what it really says. Every mathematical statement of experi- mental relationships in physics is no more than a statement of a relationship between numbers. By m i and m 2 we simply mean the numerical measures of the masses of particles 1 and 2 in terms of some arbitrarily chosen unit. The concept of mass is one that has been developed to aid us in our description of nature— like- wise the concepts of force and distance. But it is always the numbers that we deal with. Thus the full verbal equivalent of Eq. (5-1) would be: The numerical measure of the force F with which any particle at- tracts any other is proportional to the product of the numerical measures of their masses and inversely proportional to the square of the numerical measure of their separation. Keep those italicized phrases in mind whenever you read the mathematical statement of a physical relationship. They are almost never included explicitly, yet without them there is a danger of reading more into such a statement than it really con- tains. Thus the customary type of abbreviated colloquial state- ment of the law of gravitation begins: "The force of gravitational attraction between two particles is proportional to the product of their masses . . . ." What, one may ask, is meant by the product of two masses? What sort of a physical quantity is that? Even very good seientists have been drawn, at times, into almost metaphysical arguments through the effort to read some special significance into the "dimensions" of such combinations. By reminding oneself what an equation such as Eq. (5-1) actually represents, one can avoid any such confusion. The classic experiment to measure G was performed by the British physicist Henry Cavendish in 1798. It involved a 141 Gravitational forees Fig. 5-1 Sehemat ic diagram of a gravity lorsion- balance experiment. measurement of the gravitational foree between solid lead spheres of modest dimensions and made use of an ingenious mechanical arrangement — a torsion balance — to detect this tiny attractive foree. Figure 5-1 shows the essential features of the arrangement. Two small lead spheres are placed at the ends of a light rod. The rod is suspended horizontally by a very thin metal fiber attached to its midpoint. Two larger lead spheres are positioned elose to the smaller ones, in symmetrical fashion, so that the gravita- tional attraction between these pairs of large and small masses tends to rotate the rod in a horizontal plane. An equilibrium orientation is reached when the twist of the supporting fiber provides a restoring effect that just balances the gravitational attraction. A beam of light reflected from a small mirror attached to the rod allows the tiny angular deflection to be amplified into a substantial movement of a spot formed by the reflected light on a distant wali (the "optical lever" effect). A torsion-balance arrangement of this type, incorporating an optical lever, is one of the most sensitive of all mechanical devices. The gravitational foree is astonishingly weak under the conditions of a laboratory cxperiment involving the interaetion of relatively small objects, and the detection and measurement of it is an extremely delicate operation. For example, a modern version of the Cavendish apparatus 1 uses two small suspended lead spheres, each of 15-g mass, and large spheres of 1.5 kg each. The center-to-center distance between a small sphere and a ■Manufactured by the Leybold Co. 142 The va-rious forees of nature Fig. 5-2 Globular cluster ofstars held together by graoita- lionalforces (globular cluster M 13 in the constellation Her- cules). (Photograph from the Hale Obser- vatories.) large one is about 5 cm. Under these conditions the gravita- tional force of attraction is only about 6 X 10~ 10 N. The weight of a single human hair is about 10,000 times greater than this! Although the gravitational interaction is intrinsically so very feeble, it plays the prime role in most astronomical systems, because (1) the interacting objects are extremely massive and (2) other forces are almost absent. An interesting example is pro- vided by globular star clusters. These are collections of stars in a spherically symmetric distribution ; more than 120 such clusters have been identified in our galaxy. Figure 5-2 shows a globular cluster containing perhaps more than a million stars. One can infer from the symmetry of this system that there is probably no net rotational motion to the cluster as a whole though individual stars undoubtedly travel in all directions. Direct experimental evidence on the actual motions of individual stars is very meager. If we consider a star "at rest" near the outer edge of the cluster, the net force due to all other stars should be (by symmetry) toward the center of the cluster and the star will accelerate, reaching its maximum speed at the center. The net force on the star (again by symmetry considerations) will ap- proach zero as the star reaches the center of the cluster. Having gained considerable speed, the star will pass through the center and gradually slow down due to the resultant force of attraction 143 Gravitational forces Fig. 5-3 Schematic diagram ofa group of globular clusters asso- ciated with our Galaxy. of all the other stars (which at all times is a net force directed toward the center of the cluster). The star will finally reach a point diametrically opposite its starting position. Thus a star could perform oscillations along a diameter passing through the center of cluster. Because the separation between stars is so very much greater than the diameter of any star, the chance of a stellar collision is very small, even at the center of the cluster (although the photograph would not suggest this). Thcre is clear astronomical evidence for other systems of a similar type but on a much larger scale. Our Galaxy is surrounded by a spherically symmetric "halo" of globular clusters (see the sketch shown in Fig. 5-3). Since the individual clusters maintain their compact identity, each can itself be considered a "mass point," and together they form a sort of supercluster of globular clusters. Direct observational evidence shows that clusters are traveling in all directions and there is no reason to doubt that a single cluster can perform oscillations through the heart of our Galaxy (just as single stars can oscillatc within a cluster), again with almost negligible chance of individual stars colliding with each other. These two similar examplcs illustrate cases where gravita- tional attraction between masses is the sole force that governs the motion. Usually gravitational forces are important only where at least one body of astronomical size is involved — note that it is solely because the earth is in this category that gravity exercises a major influence on our everyday lives. One can think of certain exceptions to this general rule, as for example the initial stages of the aggregation of neutral hydrogen atoms 144 The various forces of naturc under their mutual gravitational attractions to form a proto- galaxy. ELECTRIC AND MAGNETIC FORCES The forces that electrically charged particles exert on one another are of fundamental importance in nature. The basic law of electric force is that found by the nineteenth-century French physicist C. A. Coulomb, and known by his name. Like Cavendish, Coulomb used a torsion balance for his measure- ments. 1 But whereas Cavendish simply measured the gravita- tional constant, taking the basic form of the force law as already known, Coulomb explored the actual form of the law of electric force. Coulomb's law states that a charged particle at rest will attract or repel another charged particle at rest with a force proportional to the product of the charges, inversely propor- tional to the square of their separation and directed along the line separating the two particles. 2 The force is attractive when the charges are unlike and repulsive when they are alike in sign. If one denotes by qi and q 2 the charges carried by the particles, the magnitude of the force that particle 1 exerts on particle 2 is given by *»-km (5-2) identical in form with the law of universal gravitation. The metric unit of electric charge is the coulomb (C). The constant k in Coulomb's law has the value 9 X 10 9 N-m 2 /C 2 . A coulomb is a huge amount of electric charge — vastly more than is usually found in isolation in nature. One coulomb separated from a similar charge 1 mile distant would experience a repulsive force of about 3500 N (approximately 800 lb). Yet an electrically neutral droplet of water | in. in diameter contains nearly 1000 C of positive charge in the nuclei of the hydrogen and oxygen atoms, balanced by an equal amount of negatively charged electrons. (Check this for yourself.) 'Their experiments were performed at almost the same time, but apparently they acted quite independently of one another in their choice and development of the torsion-balance technique. 2 Note that insertion, where appropriate, of the phrase "numerical measure of" is tacitly assumed. 145 Electric and magnetie forces Fig. 5-4 Cakulaled irajectories ofprotons ofaboul 14 Gev kinetic energy approaching the earth 'm its magnetic egualorial place and deflected by its magnetic field. (After D. J. X. Monlgomery, Cosmic Ray Physics, Princeton Unioersity Press, 1949.) It is interesting to compare the sizes of electrical forces with those of gravity. Consider, for example, the gravitational torsion balance mentioned earlier. If only one out of every 10 ' 8 electrons in each lead sphere were missing, the resultant imbalance of electric charge on the masses would produce an electrical force comparable to the gravitational force. Through such examples one can appreciate the immense strength of the electrical force compared to the gravitational force. Although gravity is always present, the electrical force is overwhelmingly the most significant agent in all chemical and biological processes and in the interactions between physical objects of everyday size (i.e., below those of the astronomical domain). It holds atoms together, provides the rigidity and tensile strength of material objects, and is the only force involved in chemical reactions. We have been discussing the electrical forces between stationary charged particles. Moving charged particles also exert electrical forces on each other. But an additional force arises in this case which we call the magnetic force. It has the interesting property that it depends on the velocity of the charges and always acts on a given charged particle at right angles to the particle's motion. We shall discuss these magnetic forces more precisely in Chapter 7. An illustration of the magnetic force is provided by the trajectories of protons (positively charged) ejected from the sun and approaching the earth. Figure 5-4 shows possible paths of protons approaching in the equatorial plane, when they pass 146 The various forces of nature in the vicinity of the earth's magnetic field. It is confidently believed that the earth's magnetic field itself arises as a result of charged particles in motion inside the earth, so basically this is an example of the electromagnetic interaction between charges in motion. Actually, from the standpoint of relativity theory, the mag- netic force is not something new and different. Charges that are moving with respect to one observer can be stationary with respect to another. Thus, if one accepts the basic idea of rela- tivity, one may expect to be able to relate a magnetic force, as observed in one reference frame, to a Coulomb force, as ob- served in another frame. For the detailed working out of this idea, see for example the volume Special Relativity in this series. NUCLEAR FORCES Although electric forces are responsible for holding atoms to- gether, they would, by themselves, prevent the existence of atomic nuclei. For we all know that nuclei contain protons, electrically repelling one another and not stabilized by a com- pensating negative charge. But nature has supplied another force, known as the sirong interaction, which binds together the nucleons (protons and neutrons) in a nucleus. Although much stronger than the Coulomb force at sufficiently short distances, its properties were relatively unknown until recently because of its extremely short range. For distances greater than about 10 -13 cm (= 1 F) this nuclear force quickly becomes negligibly small, but it dominates over all other interactions between nucleons at shorter distances. It is an exceedingly complex type of interaction, attractive down to about 0.4 F and strongly re- pulsive for still smaller separations. It is, in part, a noncentral force— that is, in contrast to the gravitational and Coulomb interactions, it is not directed along the line joining the centers of the interacting particles. Somewhat analogously to the Cou- lomb force, which exists only between electrically charged particles, the strong nuclear interaction exists only among a certain class of particles, known as hadrons (Gr. hadros: heavy, bulky), which besides the nucleons themselves includes a number of lighter particles (mesons) and heavier particles (baryons), all of which are unstable and very short-lived (10 -8 sec or less). Another type of force associated with nuclear interactions 147 Nuclear forces is also known to exist but is not very well understood as yet. It is called the "weak interaction." The range of this force is even less than that of the strong interactions, and its strength is estimated to be only about 10 -15 as great as the other. This weak interaction is important only for certain types of nuclear process, such as radioactive beta decay. It is instructive to compare the magnitudes of the different types of forces exerted between two protons 10~ 15 m apart — i.e., separated by about one nucleon diameter: Type of interaction Approximate magnitude of the force, N Gravitational 2 X 10-" Coulomb 2 X 10 2 Nuclear (strong) 2 X 10» The magnitude of the weak nuclear interaction, within the range in which it is effective, is of the order of 10~ 1 3 of the electrical force, but even so it is greater, by a factor of the order of 10 23 , than the calculated gravitational force at the same distance be- tween two particles. The nuclear forces have really no place in a discussion of Newtonian mechanics. Indeed, the very use of the word "force" in connection with nuclear interactions is open to question, for any statement about the force between two nuclear particles is at best a remote inference. We cannot cite any direct observa- tions comparable to those of Cavendish and Coulomb for gravitational and electrical forces. Moreover, the subject of nuclear forces, like everything else in subatomic physics, requires the ideas and techniques of quantum mechanics, which is a theory that has almost no use for the concept of force as such. Thus, when we talk of classical or Newtonian mechanics, we are really concerned with situations in which the only relevant kinds of interactions are electromagnetic or gravitational. FORCES BETWEEN NEUTRAL ATOMS It is a fact of profound importance that, in contrast to gravita- tion, the electric forces between individual particles may be repulsive as well as attractive, because of the existence of two different signs of electric charge. This duality of charge makes possible the existence of mattcr that is clectrically neutral in 148 The various forces of nature bulk and of atoms that are individually neutral by virtue of having equal numbers of protons and electrons. 1 On the face of it, therefore, two neutral atoms would exert no forces on one another at all when separated (if one excepts the usually negligible gravitational force). This, however, is not quite true. It would be the case if the positive and negative charges in an atom were located at a single point. We know, however, that the electrons and the nucleus are separated by a certain small amount. Also, the atom is not a rigid structure; and although the "center of gravity" of the negative charge distribution due to the electrons coincides with the positive nucleus when the atom is isolated, the approach of another particle can disturb this situation. One manifestation of this is a characteristic force of attraction between neutral atoms— a force named after the great Dutch physicist J. van der Waals. This force increases much more rapidly than l/r 2 as two atoms ap- proach one another— or (to put it in a way that carries a subtly different message) the force falls off with increasing distance much more rapidly than the attraction between two unbalanced charges of opposite sign. The basis of the van der Waals force is still, of course, the inverse-square law of force between point charges, but its detailed character cannot be calculated without the use of quantum mechanics. The final result is a force varying as l/r 7 between neutral atoms of the same kind. There is, however, another kind of force that comes into play between neutral atoms when the attempt is made to squash them together. This is a positive (repulsive) force that increases with decreasing separation even more rapidly than the van der Waals force. The result is that the net force exerted by one neutral atom on another, as a function of the separation r be- tween their centers, has the kind of variation shown in Fig. 5-5. The force passes through zero at a certain value of r, and this value (r o) can be thought of as the sum of the two atomic radii— i.e., one atomic diameter if the atoms are identical. The repulsive component of F grows so sharply with further decrease of r below r that atoms behave to some approximation as hard spheres; one manifestation of this is the highly incompressible 'J. G. King has shown by direct invesligation of the degree of neutrality of a volume of gas that the basic units of positive and negative charge in ordinary matter cannot differ by more than about 1 part in 10" (private communica- tion). 149 Forces between neutral atoms Fig. 5-5 Qualitative graph oflhe force between two neutral aloms as a function of the distance be- tween their centers. The dashed line represents the nonphysical idealization of aloms t hai act as completely hard spheres Ihal attract one anolher. nature of condensed matter. The dashed line in Fig. 5-5 in- dicates the result of idealizing the interatomic force to correspond to complete impenetrability for r < r and the van der Waals attraction for r > r . This model can be used quite effectively to analyze the deviations of an actual gas from the ideal gas laws. Since almost all the objects we deal with in classical me- chanics are electrically neutral, this basic atomic interaction — the electric force between neutral particles — is of fundamental importance, as we shall discuss next. CONTACT FORCES Many of the physical systems we shall deal with are the ordinary objects of everyday experience, acted upon by such forces as friction, the push and puli of struts and beams, the tension of strings and cables, and so on. Each of these forces involves what we naively call physical "contact" with the object under ob- servation. Consider, for example, a book resting on a horizontal table. The book is supported by the sum total of countless electro- magnetic interactions between atoms in the adjacent surface layers of the book and the table. A submicroscopic analysis of these interactions would be prohibitively complex. For most purposes, however, we can ignore this complexity and can lump all these interactions together into a single force that we shall call a contact force. This is a rather artificial category but a useful one. Broadly speaking, all the familiar forces of a me- chanical nature, including the force that a liquid or a gas exerts on a surface, are contact forces in this sense. Our discussion in the previous section makes it clear that they are electric forces 150 The various forces of nature Fig. 5-6 (a) Two charged spheres, obciously not in conlact. The weight of the upper sphere is balanced by the electrical repulsion. (b) Two uncharged spheres, apparently in conlact. Oli a sufficiently magnified scale, this appearance ofa conlact ofsharp geometrical bound- aries would disappear. Fig. 5-7 (a) Quali- tatwe graph offorce versus separation for the charged spheres ofFig. 5-6{a). The conlact is "soft" — i. e., the difference be- tween F and W varies slowly with r. (b) Comparable graph for the uncharged spheres ofFig. 5-5(6). The conlact is "hard" — the near equality of F and W occurs only over an extremely narrow range of values of r. exerted between electrically neutral objects. The development of such forces when one smooth, hard object is pressed against another comes about through a distortion of the distributions of positive and negative electric charges. It is characteristic of such forces that their variation with the distance between the objects is much more rapid than the inverse-square dependence that holds for objects carrying a net charge. Thus the contact forces are, in effect, forces of very short range; they fail to negligible size when objects are more than about one atomic diameter apart. The fact, however, that they do have a syste- matic dependence on the distance of separation means that the notion of what we ultimately mean by "contact" is not clear-cut. There is no fundamental distinction between the situation repre- sented by two charged spheres, visibly held apart by their elec- trical repulsion, and the same two spheres, uncharged, apparently in contact, as shown in Fig. 5-6. In each case the upper sphere assumes a final position of equilibrium in which the net elec- trical force on it just balances its weight. For any smaller separation it exceeds the weight. But if we display these variations of force with distance, as in Fig. 5-7, we see a drastic difference. In the case of uncharged objects, the transition from negligible force to very large force is so abrupt that it supports our im- pression of a completely rigid object with a geometrically sharp Separation of centers, r (a) Separation of centers. t (b) H 151 Contact forces boundary and no interaction outside that boundary. Do not forget, however, that this is an idealization, and that sufficiently refined measurements would always reveal a continuous variation of force with separation. FRICTIONAL CONTACT FORCES Fig.5-8 (a) Block on a rough horizontal table, subjected to a horizontal puli, P. {b) Qualitative graph of the frictional force, S, as afunclion ofP. The condiiion S = P can be satisfied up to the point at which S = nN. After that, the eguilibrium is bound to be broken. The discussion in the previous section has concentrated on contact forces called into play by the simple pushing together of two objects. Such forces are then at right angles to the surface of contact — we call them normal forces in the geometrical sense of that word. But much importance and interest attaches to the tangential forces of friction that appear when the attempt is made to drag an object sidcways along a surface. Figure 5-8(a) depicts a block resting on a horizontal surface; its weight is supported by a normal contact force N. We now apply a hori- zontal force P. Suppose that the magnitude of P is gradually increased from zero. At first nothing seems to happen ; the block remains still. We know, from our analysis of equilibrium situa- tions, that this means that a force equal and opposite to P is being supplied via the contact between the block and the surface. This is the frictional force, S. It automatically adjusts itself to balance P, just as the normal force N would automatically in- crease if we deliberately pushed down harder, in a vertical di- rection, on the top of the block. In both cases we can imagine minute deformations of the electric charge distributions along the interface, sufficient to develop the requisite forces. But then, when P is increased beyond a certain value, the frictional force S is no longer ablc to keep step with it. The equilibrium is broken down, and motion ensues. A graph of S against the applied force P might look like Fig. 5-8(b). Once S has been brought to its maximum limiting value, it may even drop at first as P is 152 The various forces of naturc Fig. 5-9 (a) Force on a sphere in a flowing fluid. (b) Total force o f fluid friction is mode 1 1 P ofseparate terms that are respectively linear and quadratic in the relativeflow celocity, v. F v «(i')/ b mtr-i -« — r^ ) A v ^ — O further increased, although it tends to remain at a fairly constant value thereafter. However, the whole regime P > S corresponds to motion, and ff may depend in detail on the velocity. The only uniquely defined region is that of static equilibrium, throughout which we can put JF = P, as represented by the 45° line on the graph of Fig. 5-8(b). The other feature of interest is the empirical fact that the limiting value of £F is roughly proportional to the normal force N, so that their quotient — the coefficient of friction (n) — is a property of the two surfaces in contact: = 11N (5-3) The above discussion applies to the contact of two solid surfaces. If the contact is lubricated, the behavior is very dif- ferent. One is then dealing, in effect, with the properties of the contact between a fluid (liquid or gas) and a solid. The basic properties of so-called fluid friction can be studied by measuring the force exerted on a fixed solid object as a stream of fluid is driven past it at a given speed d [see Fig. 5-9(a)]. Over a wide range of values of u, this fluid frictional resistance is well described by the following formula: R(v) = Ao + Bv 2 (5-4) where A and B are constants for a given object in a given fluid. The first term depends on the viscosity of the fluid, and the second is associated with the production of turbulence. Since the ratio of the second term to the first is proportional to v, one knows that at sufficiently high speeds the fluid friction is domi- 153 Frictional contaci forces nated by turbulence, however small the ratio B/ 'A may be. The same consideration guarantees that at sufficiently low speeds the resistance will be dominated by the viscous term, directly propor- tional to o [see Fig. 5-9(b)]. CONCLUDING REMARKS PROBLEMS In this chapter we have given a brief account of the three major types of physical interactions and have indicated the general areas in which they are dominant. To recapitulate: Nuclear forces are significant only for nuclear distances, the gravitational force is important only if objects of astronomical scale are in- volved, and nearly everything else ultimately depends on electro- magnetic interactions. The study of physics is essentially the attempt to understand these interactions and all their con- sequences. In mechanics we have, for the most part, the more modest goal of taking the forces as given and considering various dynamical situations in which they enter. We shall, however, be discussing two classic cases— gravitation and alpha-particle scattering — in which Newtonian mechanics provided the key to the basic laws of force. The present chapter has provided a kind of preview, because it summarizes the state of our current knowl- edge without entering into any detailed discussion of how we have come to know it. The real work lies ahead! 5-1 At what distance from the earth, on the line from the earth to the sun, do the gravitational forces exerted on a mass by the earth and the sun become equal and opposite? Compare the result with the radius of the moon's orbit around the earth. 5-2 By what angle, in seconds of are, will a plumbline be pulled out of its normal vertical direetion by the gravitational attraction of a 10-ton truck that parks 20 ft away? Do you think that this effect could be detected ? 5-3 In a Cavendish-type apparatus (see the figure) the large spheres are each 2 kg, the small spheres each 20 g. The length of the arm con- necting the small spheres is 20 cm, and the distance between the centers of a small sphere and the big sphere elose to it is 5 cm. The torsion constant of the suspending fiber is 5 X 10 -8 m-N/rad. The angular deflection of the suspended system is deduced from the displacement 154 The various forces of naturc Scale ^20 g of a reflected spot of light on a scale 5 m away. (Remember that the change in direction of the reflected light is twice the angle through which the mirror is turned.) It is observed that when the large spheres are moved from their initial positions to equivalent positions on the other side (dashed lines) the mean position of the spot of light is shifted by 8 cm. (a) Deduce the value of G according to these data, ignoring the effect on each small sphere of the force due to the more distant of the larger spheres. (b) Estimate the percentage correction on the result of (a) re- quired to allow for the effect of the more distant spheres. 5-4 The original Cavendish experiment was done with a large-size apparatus, as is natural if one wants to make the gravitational forces and torques as big as possible. However, this requires a strong, stiff wire to support the suspended masses. Much later (1895) C. V. Boys made a miniature apparatus, using thin fibers of fused quartz for the suspensions. It is an interesting exercise to see how the attainable sensitivity of the apparatus depends on its size. Imagine two versions of the Cavendish apparatus, A and B, both using solid lead spheres, in which the radii and separations of all the masses in B, together with the length of the torsion fiber, are scaled down by a certain factor L with respect to A. We then design for maximum sensitivity in each apparatus by using the thinnest possible torsion fiber that will take the weight of the suspended masses without breaking. Now for a torsion fiber of given material and of circular cross section, the maxi- mum supportable load is proportional to d 2 , where d is its diameter, and its torsion constant is proportional to d* /I, where / is its length. Using this information, compare the maximum angular deflections obtainable with the two different sizes of apparatus. (Remember, the lengths of the torsion fibers also differ by the scaling factor L.) 5-5 The radius of the hydrogen atom according to the original Bohr 155 Problems theory is 0.5 A. (a) What is the Coulomb force between the proton and the electron at this distance? What is the gravitational force? (b) How far apart must the proton and electron be for the Coulomb force to be equal to the value that the gravitational attrac- tion has at 0.5 A? What familiar astronomical distance is this com- parable to? 5-6 Suppose electrons could be added to earth and moon until the Coulomb repulsion thus devcloped was of just the size to balance the gravitational attraction. What would be the smallest total mass of electrons that would achieve this? 5-7 For a person living at 45° latitude, what is the approximate fractional difference, during the day, between the maximum and minimum gravitational forces due to the moon — the change resulting from the fact that the earth's rotation causes the person's distance from the moon to vary ? What is a manifestation of this kind of force effect in nature? 5-8 You know that the Coulomb force and the gravitational force both obey an inverse-square law. Suppose that it were put to you that the origin of the gravitational force is a minute difference be- tween the natural unit of positive charge, as carried by a proton, and the natural unit of negative charge, as carried by an electron. Thus "neutral" matter, containing equal numbers of protons and electrons, would not be quite neutral in fact. (a) What fractional difference between the positive and negative elementary charges would lead to "gravitational" forces of the right magnitudo between lumps of ordinary "neutral" matter ? How could such a difference be looked for by laboratory experiments? (b) Is the theory tenable? 5-9 (a) The text (p. 148) quotes a valuc of the nuclear force for two nucleons close together but also suggests that to describe the nuclear interactions in terms of forces is not very practical. Can you suggest any way in which a nuclear force as such could be measured? (b) According to one of the earliest and simplest theoretical descriptions of the nuclear interaction (by H. Yukawa) the force of attraction between two nucleons at large separation would be given by F( ,) = _d e -'"o where the distance r o is about 10~ 15 m and the constant A is about 10" ' 1 N-m. At about what separation between a proton and a neutron would the nuclear force be equal to the gravitational force between these two particlcs? 5-10 Can you think of any systems or processes in which gravita- 156 The yarious forces of nature -3in.- Waterfilm lg tional, electromagnetic, and nuclear forces all play an important role? 5-11 As mentioned in the text, the attractive (long-range) part of the force between neutral molecules varies as l/r 7 . For a number of molecules, the order of magnitude of this van der Waals force is well represented by the equation F vw (r)~ -10- 76 A 7 where F V w is in newtons and r in meters. Compare the magnitude of the van der Waals force with the Coulomb force between two ele- mentary charges [Eq. (5-2), with<?i = ?2 = e = 1.6 X 10~ C]: (a) For r = 4 A. (This is a distance about equal to the diameter of a molecule of oxygen or nitrogen and hence barely exceeding the closest approach of the centers of two such molecules in a collision.) (b) For a value of r corresponding to the mean distance between molecules in a gas at STP. 5-12 One of the seemingly weakest forms of contact force is the surface tension of a liquid film. One of the seemingly strongest is the tensile force of a stretched metal wire. However, when expressed in terms of a force between individual atoms in contact, they do not look so different. Use the following data to evaluate them in these terms: (a) If a water film is formed between a rectangular wire frame, 3 in. wide, and a freely sliding transverse wire (see the figure), it takes the weight of about 1 g to prevent the film from contracting. This contractile force can be ascribed to the contact of the atoms lying within a monomolecular layer along each side of the film. Supposing that the molecules are 3 A across and closely packed, calculate the force per molecule. (b) A copper wire of 0.025-in. diameter was found to break when a weight of about 10 kg was hung from its lower end. First, calculate this breaking force in tons per square inch. If the fracture is assumed to involve the rupturing of the contacts between the atoms on the upper and lower sides of a horizontal section right across the wire, calculate the force per atom, assuming an atomic diameter of about 3 A. 5-13 A time-honored trick method for approximately locating the midpoint of a long uniform rod or bar is to support it horizontally at any two arbitrary points on one's index fingers and then move the fingers together. (Of course, just finding its balance point on one finger alone works very well, too!) Explain the workings of the trick method, using your knowledge of the basic principles of static equi- librium and a property of frictional forces: that they have a maximum value equal to a constant /x (the coefficient of friction) times the com- ponent of force normal to the surface of contact between two objects. 157 Problems 5-14 (a) A string in tension is in contact with a circular rod (radius r) over an arc subtending a small angle A0 (see the figure). Show that the force with which the string presses radially inward on the pulley (and hence the normal force AN with which the pulley pushes on the string) is equal to TA9. (b) Hence show that the normal force per unit length is equal to T/r. This is a sort of pressure which, for a given value of T, gets bigger as r decreases. (This helps to explain why, when a string is tightly tied around a package, it cuts into the package most deeply as it passes around corners, where r is least.) (c) If the contact is not perfectly smooth, the values of the tension at the two ends of the arc can differ by a certain amount AF before slipping occurs. The value of AT is equal to y. AN, where n is the coeffkient of friction between string and rod. Deduce from this the exponential relation T(6) = T e p» where To is the tension applied at one end of an arbitrary arc (0) of string and T(6) is the tension at the other end. (d) The above result expresses the possibility of withstanding a large tension T in a rope by wrapping the rope around a cylinder, a phenomenon that has been exploited since time immemorial by sailors. Suppose, for example, that the value of n in the contact between a rope and a bollard on a dock is 0,2. For 7" = 100 lb, calculate the values of T corresponding to one, two, three and four complete turns of rope around the bollard. (It is interesting to note that T is proportional to To- This allows sailors to produce a big puli or not, at will, by having a rope passing around a continuously rotating motor-driven drum. The arrangement can be described as a force amplifier.) 5-15 In a very dclicate torsion-balance experiment, such as the Cavendish experiment, the stray forces due to the fluid friction of slow air currents pushing on the suspended system may be quite significant. To make this quantitative, consider the gravity torsion-balance ex- periment described in Problem 5-3. For suspended spheres of the size stated (r « 0.8 cm), the force due to a flow of air of speed v is given approximately by the formula [Eg. (5-4)] -s,,2 R (newtons) = 2.5 X 10- 6 <; + 5 X 10 where u is in m/sec. Calculate the value of v that would cause a force due to air currents that equaled the gravitational force exerted on the sphere in this experiment (i.e., the force exerted on a 20-g sphere by a 2-kg sphere with their centcrs 5 cm apart). (Hint: Do not bother to 158 The various forces of nature solve a quadratic equation for v. Just find the values of d for which the contributions to R, taken separately, would equal the gravitational force. The smaller of the two values of u so obtained is clearly already enough to spoil the experiment.) 159 Probiems To tell us that every species of things is endowed with an occult specific quality by which it acts and produces manifest effects, is to tell us nothing. But to derive two or three general principles of motion from phenomena, and afterwards to tell us how the properties and actions of all corporeal things f ollow from those manifest principles, would be a very great step in Philosophy, t/wugh the causes of those principles were not yet discovered. newton, Opticks (1730) Force, inertia, and motion THE PRINCIPLE OF INERTIA the preceding chapters have treated matter, motion, and force as separate topics. Now we come to the central problem of Newtonian dynamics: How are motions of material objects affected by forces? We shall preface this with a question that is, on the face of it, much simpler: What can we say about the motion of an object that is subjected to no forces? It was in the analysis of this problem by Galileo that the science of dynamics really began. ] We saw in Chapter 4 how the study of static cquilibrium situations leads us to a basic principle: For an object at rest, the net force aeting is zero. What could be more natural than to turn this statement around and infer that, if the net force on an object is zero, the object must remain at rest — and, as a kind of corollary to this, to conclude that in order to keep an object moving a net force must bc maintained. After all, our experience >You will undoubtedly have solved many problems in the use of Newton's laws before rcading this chapter. Do not, on that account, assume that the following diseussion is superfluous. A wish to get down to business — writing equations and using them — is very sound. The quantitative use of a physical theory is an essential part of the game; physics is not a spectator sport. But to gain real insight and understanding — Where do the equations come from? What do they really say? — one must also examine the basic assumptions and phenomena. And some of the greatest advances in physics have come about in just this way. Einstein arrived at special relativity by thinking deeply about the nature of time. And Newton, when asked once about how he gained his insight into the problems of nature, replied "By constantly thinking unto them." 161 Fig. 6-1 Limitalions on in- ertial molion at the earth's surface. An object starling oul horizontally in a Iruly straight line at A mighl end up on a hilttop at B. In the process it would be bound to slow down. shows that moving objects on the earth's surface do come to rest if left to themselves, and a continuing effort does have to be applied to keep an object moving steadily. But, as Galileo was the first to realize, an extrapolation beyond the range of ordinary experience is possible; it is expressed in his principle of inertia. Initially, this simply asserted that an object would, if free of all resistive forces, continue with unchanging speed on a horizontal plane. Galileo himself recognized that this assertion is true only in a limited sense. For a truly flat horizontal plane is tangent to the earth's surface; therefore, if extended far enough, it must be seen as going perceptibly up hill (Fig. 6-1), and objects traveling outward along it must ultimately slow down. ' Subsequently Isaac Newton stated the principle of inertia in a generalized form in his "first law" of motion as presented in the Principia: "Every body perseveres in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed upon it." It is a familiar statement, which we probably all learn in our first encounter with me- chanics, and Fig. 6-2 shows a practical illustration of it. But what does it really say? The first thing we must recognize is that, as discussed in Chapter 2, every statement about the motion of a given object involves a physical frame of reference; we can only measure displacements and velocities with respect to other objects. Thus the principle of inertia is not just a clear-cut statement about the behavior of individual objects; it goes much deeper than that. We can, in faet, turn it around and make a statement that goes roughly as follows: There exist certain frames of reference with respect to which the motion of an object, free of all external forces, is a motion in a straight line at constant velocity (including zero). ■For Galileo's own diseussion of these matters, see his Dialogues Concerning Two New Sciences (H. Crew and A. de Salvio, translators), Dover Publica- tions, New York. 162 Force, inertia, and motion Fig. 6-2 A motion al conslanl velocity — a bullet traveling at about 1,500 J 1 /sec, photographed strobo- scopically at 30,000 flashes/sec. (Photograph by Prof. Harold E. Edgerton, M. I. T.) A reference frame in which the law of inertia holds good is called an inertial frame, and the question as to whether a given frame of reference is inertial then becomes a matter for observation and experiment. Most observations made within the confines of a laboratory on the earth's surface suggest that a frame of reference attached to that laboratory is suitable. After all, it was on the strength of observations within such a frame that Galileo arrived at the principle of inertia in the first place! A more critical seruti ny shows that this is not quite good enough, and we need to look further afield — but we shall do that later. For the moment we shall limit ourselves to introducing the main principle, which is not affected by the later refinements. Sir Arthur Eddington, who had a flair for making comments that were both penetrating and witty, offered his own version of the principle of inertia: "Every body continues in its state of rest or uniform motion in a straight line, except insofar as it doesn't." 1 In other words, he regarded it as being, in the last 'A. S. Eddington, The Nalure o/ the Physical World (Ann Arbor Paperbacks), University of Michigan Press, Ann Arbor, Mich., 1958. 163 The principle of inertia analysis, an expendable proposition. (He was paving the way for a discussion of general relativity and gravitation.) This remark of Eddington's nevertheless draws attention, in a colorful way, to what is the very foundation of Newtonian mechanics: Any deviation from a straight-line path is taken to imply the existence of a force. No deviation, no force — and vice versa. It must be recognized that we cannot "prove" the principle of inertia by an experimental test, because we can never be sure that the object under test is truly free of all external interactions, such as those due to extremely massive objects at very large distances. More- over, there is the far from trivial question of defining a straight line in a real physical sense: it is certainly not intuitively obvious, nor is it an abstract mathematical question. (How would you define a straight line for this purpose?) Nevertheless, it can be claimed that the principle of inertia is a valid generalization from experience; it is a possible interpretation of observed motions, and our belief in its validity grows with the number of phenomena one can correlate successfully with its help. FORCE AND INERTIAL MASS: NEWTON'S LAW The law of inertia implies that the "natural" state of motion of an object is a state of constant velocity. Closely linked to this is the recognition that the effect of an interaction between an object and an external physical system is to change the state of motion. For example, we havc no doubt that the motion of a tennis ball is affected by the racket, that the motion of a compass needle is affected by a magnet, and that the motion of the earth is affected by the sun. "Inertial mass" is the technical phrase for that property which determines how difficult it is for a given applied force to change the state of motion of an object. Let us consider how this description of things can be made quantitative. "Force" is an abstract term, but we have seen how to asso- ciate it with practical operations such as compressing springs, stretching rubber bands, and so on. We can readily study the effect of pushing or pulling on an object by such means, using forces of definite magnitude. The observations become par- ticularly clear-cut if we apply a force to an object that would otherwise move with constant velocity. A very close approxima- tion to this ideal can be obtained by supporting a flat-bottomed object on a cushion of gas— for example by placing the object on 164 Force, inertia, and motion I 3 I 4 I © (a) Fig. 6-3 (a) Strobo- scopic pholograpli of a uniformly accel- eraled molion. The time interval between Hght flashes was £ sec. (From PSSC Physics, D. C. Heath, Lexington, Massa- chuselts, 1960.) (b) Simple dynamical experimenls ihat can be used as a basis for deoeloping Newton' s second law. 2 springs o 1 block 1 block 1 spring Acceleratior (b) Acceleration = « a ° a horizontal table pierced with holes through which air is blown from below. It is then possible to puli horizontally on the object and make such observations as the following (see also Fig. 6-3): 1. A spring, stretched by a constant amount, causes the velocity of the object to change linearly with time — the accelera- tion produced in a given object by a given force is constant. 2. If a second spring, identical with the first and stretch- by the same amount, is used side by side with the first, the ac- celeration is doubled. That is, if we take a known multiple of a force, according to our criteria for comparing forccs in static equilibrium (see Chapter 4), then the acceleration produced in a given object is directly proportional to the total force. It thus becomes possible to write down simple equations expressing the relation betwccn forces F and accelerations a: 165 Force and ineriial mass: Newton's law a = kF or F= k'a where k and k' describc thc inertial properties of the particular object. Which of the above two statements of the results is more convenient? We find the answer in another simple experiment: 3. If we place on the first object a second, identical object, it is observed that all accelerations produced by given arrange- ments of the springs are rcduced to half of what was obtained with one object alone. We can express this most easily by choos- ing the second of the above equations, so that the inertial property is additive — i.e., the inertial constants k' of two different objects can be simply added together, and the acceleration of the com- bined system under a given force is immediately given by F = (Ar'i + k' 2 )a l+2 i.e., 1+2 fci + JG It is by such steps that one can be led to the equation that is universally known as "Newton's (second) law": F = ma ■= m— (6-1 ) di where the proportionality factor m (identical with k' as defined above) is called the inertial mass of the object and F is the net force acting on it. Embodied in this basic statement of Newton's law is the feature that force and acceleration are vector quan- tities and that the acceleration is always in the same direction as the net force. An interesting historical fact, often overlooked, is that Newton's own statement of the basic law of mechanics was nol in the form of Eq. (6-1); the equation F = ma appears nowhere in the Principia. Instead, Newton spoke of the change of "motion" (by which he meant momentum) and related this to the value of force X time. In other words, Newton's version of the second law of motion was essentially the following: FAt = mAo (6-2) 166 Force, inertia, and motion We shall see in Chapter 9 that Newton's way of formulating the law grew, by inference, out of the particular kind of evidence available to him: the consequences of collision processes. Direct experiments of the kind illustrated in Fig. 6-3 were not possible with the limited techniques of Newton's day. (The only type of uniformly accelerated motion easily accessible to Newton was that of motion under gravity, but this, of course, did not allow any independent control over the value of F applied to a given object — unless the use of an inclined plane, giving a driving force mg sin 6, is regarded as fulfilling this purpose.) SOME COMMENTS ON NEWTON'S LAW Sitnple and familiar as Eq. (6-1) is, it nevertheless contains an enormous wealth of physical concepts — indeed, almost the whole basis of classical dynamics. First comes the assumption that quantitative measurements of displacements and time intervals lead us to a unique value of the acceleration of an object at a given instant. If we remind ourselves that displacements can only be measured with respect to other physical objects, we see that this, like the principle of inertia, cannot be separated from the choice of reference frame. In fact, we tacitly assume that the frame in which the acceleration is measured is an inertial frame. For the kinds of basic experiments illustrated in Fig. 6-3, the earth fills this role. Next comes the feature, already emphasized, that the ac- celeration vector is in the direction of the net force vector. This is an important result; it is an expression of the fact that the accelerative effects of several different forces combine in a linear way. Suppose, for example, that an object has two springs attached to it [see Fig. 6-4(a)]. Spring 1 exerts on the object a force Fi in the x direction. Acting alone it would produce an acceleration F t /m along x. Spring 2 exerts a force F 2 i n the y direction. Acting alone it would produce an acceleration F-i/m along y. It is then a matter of experiment — not predictable by pure logic — that the acceleration caused by the two springs acting together is just what one would calculate by adding the vectors Fi and F 2 to form a resultant force F [Fig. 6-4(b)] and applying this single force to the mass [Fig. 6-4(c)]. The observed acceleration a is equal to F/m. This result provides the dynamical basis for the "independence of motions" that we discusscd as a purely kinematic effect in the trajectory problem of Chapter 3. It tells us that the instantaneous acceleration of an object is the 167 Some comments on Newton's law Fig. 6-4 (a) Object pulled by two springs in perpendic- ular directions. (b) Resultan! force calculated according to Ihe laws ofveclor addUion. (c) Obserced acceleralion ofthe objecl agrees With lhal due lo the net force cector as found in (b). consequence of a linear superposilion of the applied forces or of the accelerations that they would individually produce. If this resiilt did not hold, the prediction and analysis of motions as produced by forces would become vastly more complicated and difficult. Let us add a word of explanation and caution here. The linear superposition of instantaneous components of acceleration does not mean that we can always automatically proceed to calculate, let us say, the whole course of development of the y component of an object's motion without reference to what is happening in the x direction. To take an example that we shall consider in more detail later, if a charged particle is moving in a magnetic field, the component of force in a given direction de- pends on the component of velocity perpendicular to that direc- tion. In such a case, we have to keep traek of the way in which that perpendicular velocity component changes as time goes on. In the case of an object subjected to a single force, one may be tempted to think that it is intuitively obvious that the ac- celeration is in the same direction as the force. It may be worth pointing out, therefore, that this is not in general true if high- velocity particles are involved — sufficiently fast to require the modified kinematies and dynamics of special relativity. ' Finally comes the assertion that a given force, applied to a particular object, causes the velocity of that object to change at a certain rate a, the magnitude of which depends only on the ■See, for example, the volume Special Relatioity in this series. 168 Force, inertia, and motion Fig. 6-5 Increase of i inertial mass with speed, as revealed in experiments on high- speed electrons. Based on data of (open circles) Kauf- mann (1910), (filled circles) Bucherer (1909), and (crosses) Guye and Lauanchy (1915). (AflerR.S. Shankland, Atomic and Nuclear Physics, Macmillan, New York, 1961.) magnitude and direction of F and on a single scalar quantity characteristic of the object — its inertial mass m. This is a very remarkable result; let us consider it further. Newton's law asserts that the acceleration produced by any constarit force, as for example by a stretched spring, has the same value under all conditions. Thus, according to this state- ment, it does not matter whether the object is initially stationary or is traveling at high speed. Is this alvvays, and universally true? No! It turns out that for extremely high speeds — speeds that are a significant fraction of the speed of light — the acceleration produced by a given force on a given object does depend on v. Under these high-speed conditions Newton's mechanics gives way to Einstein's, as described by the special theory of relativity: The inertia of a given object increases systematically with speed according to the formula m(o) = (1 - oVc2)" 2 (6-3) This relation is shown in Fig. 6-5, with experimental data that substantiate it. The quantity m Q , which is called the "rest mass" of the object, represents what we can simply call the inertial mass in all situations to which classical mechanics applies, because for any v « c the value of m according to Eq. (6-3) is inap- preciably different from m . Another implication of Newton's law, as expressed by Eq. (6-1), is that the basic dynamics of an object subjected to a given force does not depend on d 2 v/dt 2 or on any of the higher time derivatives of the velocity. The absence of any such com- 169 Some comments on Newton's law plication is in itself a remarkable result, which as far as we know continues to hold good even in the "relativistic" region of very high velocities. It has, however, been pointed out that if one considers physiological effects, not just the basic physics, the existence and magnitude of d 2 o/dt 2 (= da/dt) can be important. We all know the good feeling of a "smooth acceleration" in a car, and what we mean by that phrase is an acceleration that is close to being constant. A rapid rate of change of acceleration produces great discomfort, and it has even been suggested that a unit of da/dt — to be called a "jerk" — should be introduced as a quantitative measure of such effects! The conclusion that we can draw from the above discussion is that Newton's law, although ultimately limited in its applica- tion, does express with insignificantly small error the relation between the acceleration of an object and the force acting on it for almost everything outside the realm of high-speed atomic particles. SCALES OF MASS AND FORCE Granted that a given force produces a unique acceleration of a given object, we can then apply this same force to different objects. Such observations can be used to establish quantitative scales for measuring both inertial masses and forces. In taking this step, we adopt Newton's law as the central feature of me- chanics. Although we have hitherto used static situations to compare forces with one another, we now turn to dynamics for defining the absolute magnitudes of forces in terms of the motions they produce. This also means that instead of relying on static measurements to give us prior knowledge of the magnitude of a force, we accept the idea that Newton's law can be used as an analytical tool for deducing the force from the observed ac- celeration that it produces. Because the measures of force and inertial mass are linked in the single equation F = ma, there is danger of circularity in our definitions. But we shall not delve into the subtleties of this problem; we shall simply present a pragmatic method of establishing scales of measures for these quantities. Our observations permit us to assume that every time a given spring is stretched to the same elongation, it exerts the same magnitude of force on an object attached to it — for we 170 Force, inertia, and motion observe a reproducible acceleration. We can then take a number of different objects, labeled 1, 2, 3, ... , puli them one at a time with the same force (i.e., our spring stretched to the same elon- gation) and measure the individual accelerations : a x , a 2 , a 3 , We can use these experimental results to define an inertial mass scale because we can put F = m\a\ = «12^2 = W3<33 - • • Therefore, m\ fl2 W!l fl3 One particular object (e.g., m,) can be chosen (arbitrarily) as a Standard unit mass called a "kilogram." A quantitative measure of all other inertial masses can then be obtained in terms of the Standard object. Originally, the kilogram was defined to be the mass of 1000 cm 3 of water at its temperature of maximum density (about 4°C), but it is now the mass of a particular cylinder of platinum-iridium alloy, kept at the International Bureau of Weights and Measures in Sevres, France. (This object was in- tended to be exactly equivalent to the liter of water, but when a small discrepancy was discovered it was decided to switch to the metal Standard as being generally superior in terms of durability, reproducibility, and convenience.) If the inertial mass is truly a property of the object alone, then the ratios (6-4) must be independent of the particular force used. Repeating the procedure with a spring extended by a different amount and hence with different accelerations, one finds experimentally that the mass ratios are the same as before. The fact that these experimentally determined mass ratios are independent of the magnitude of the force establishes the inertial mass as a characteristic property of the object. Our quantitative scale for forces likewise stems from New- ton's law once the scale of inertial masses has been established, so that in the MKS system, as we mentioned in our first dis- cussion of forces in Chapter 4, the unit force (the newtori) is defined as the force that imparts to 1 kg an acceleration of 1 m/sec 2 : 1 N = 1 kg-m/sec 2 Dynamical units of force in other systems of measurement can 171 Scales of mass and force be defined in an exactly similar way, A newton, as we noted in Chapter 4, is about equal to the gravitational force on an apple, i.e., on a mass of a few ounces. In describing the kinds of simple experiments on which the formulation of F = ma might be based, we introduced the result that the inertial property represented by mass is additive. You may be tempted to feel that this is obvious and not worthy of an experimental test. As far as ordinary objects are concerned, this commonsense reaction to the problem is sound. But just to recognize that there is, in principle, a legitimate question here, imagine that we have measured the inertial masses of a proton and an electron, separately, and that we then let them come together to form a hydrogen atom. Is the inertial mass of the atom equal to the sum of the masses of the electron and proton? No; it is a shade less. Why? Because in the formation of the atom, with the binding together of the proton and electron, the equivalent of a tiny amount of mass escapes in the form of radiation. Conversely, if an object is made up of various parts held together by cohesive forces, so that effort has to be supplied to separate it into those parts, then the sum of the masses of the individual parts is greater than the mass of the original object. For ordinary objects the difference is immeasurably small, but in atomic and nuclear systems it can become a significant feature of the total mass, and provides the basis of calculations on the energy of nuclear reactions, etc. Returning now to the observed additivity of inertial masses for macroscopic objects, we can see that this property fully justifies the procedure of making a set of Standard masses by constructing blocks of a given material with their volumes in simple numerical ratios — e. g., 1 :2:5: 10 .... This proportion- ality of mass to volume was taken by Newton himself as basic, embodying the concept of a constant density for a given material. The very first sentence in the Principia is, in fact, a definition of mass as the product of density and volume — a definition that has drawn heavy fire, because some critics regard it as circular. "How," they say, "can one define density except as the quotient of mass and volume?" But Newton had a picture of solid matter as built up of small particles packed together in a uniform manner, and it probably seemed to him more logical to take this inner structure as primary. The calculation of the mass of a lump of matter would then be in essence a matter of counting the number of particles that it contained and multiplying by the mass of a single particle. 172 Force, inertia, and motion THE EFFECT OF A CONTINUING FORCE Our main concern in this chapter and the next is with the in- stantaneous effect of a force. Let us, however, take a first look at a question that will be the subject of very extensive analysis later in the book. This question is: What is the effect of a force that is applied to an object and maintained for a while? You are no doubt aware that the answer to this question can be given in more than one way, depending on whether we consider the time or the distance over which the force is applied. To take the simplest possible case, let us suppose that a constant force F is applied to an object of mass m that is at rest at time zero. Then we have F = ma, defining a constant acceleration. The resulting motion is thus deseribed by the most elementary versions of the kinematic equations: d = at x = \ttfi At time t the object has traveled a total distance x, and we can caleulate two possible measures of the total effect of F: Ft = mat = mv Fx = (ma)$at 2 ) = %mv 2 Thus we arrive at the two primary dynamical properties that we associate with a moving object: its momentum and its kinetic energy. The effect of F as measured by the produet Ft is called its impulse; the effect of F as measured by the produet Fx is of course what we call work. The quantitative measures of Ft and Fx are newton-seconds and newton-meters. The former, which defines our measure of momentum, does not have a special unit named in its honor. The latter, however, is expressed in terms of the basic unit of work or energy in the MKS system — the joule. Thus we have impulse — * momentum in N-sec = kg-m/sec work > kinelic energy in N-m = joules THE INVARIANCE OF NEWTON'S LAW; RELATIVITY We have emphasized how the experimental basis of Newton's law involves the observation of motions with respect to an inertial 173 The invariance o\' Newton's law; relativity Fig. 6-6 Molion of a particle P referred to iwoframes that have a relatiue uelocity v. frame of reference. The actual appearance of a given motion will vary from one such frame to another. It is worth seeing, therefore, how the dynamical conclusions are independent of the particular choice of frame — which means that Newton's me- chanics embodies a principle of relativity. The first point to establish is that, if we have identified any one inertial frame, S (i.e., a frame in which an object under no forces moves uniformly in a straight line), then any other frame, S', having a constant velocity relative to the first is also an inertial frame. This follows directly from the fact that if an object has the instantaneous velocity u in S, and if the velocity of S' relative to S is v, then the instantaneous velocity of the object relative to S' is given (see Chapter 2) by Thus if u and v are constant velocities, so also is u', and the object will obey the law of inertia as observed in S'. To discuss the problem further, let us set up rectangular coordinate systems in both frames, with their x axcs along the direction of the velocity v (see Fig. 6-6). Let the origins O and O' of the two systems be chosen to coincide at t — 0, at which instant, also, the y and z axes of S' coincide with those of S. Let a moving object be at the point P at a later time, /, when the origin O' has moved a distance vt along the x axis of S. Then the coordinates of P in the two systems are related by the follow- ing equations. (It is appropriate, in view of Galileo's pioneer work in kinematics and especially of his clear statement of the law of inertia, that they should have become known as the Galilean transformations.) 174 Force, inertia, and molion (Galilean transformalion: x' = X - - vt (o = const.) S' moves relative to S with | L' =y (6-5) a constant speed u in the z' = z -\-x direction) | r' = r The last of these equations expresses the Newtonian assumption of a universal, absolute flow of time, but it also embodies the specific convention that the zero of time is taken to be the same instant in both frames of reference, so that all the clocks in both frames agree with one another. We can then proceed to obtain relationships between the components of an instantaneous velocity as measured in the two frames. Thus for the x components we have , dx' dx ** = -d7 u ' = 'di Putting x' = x — vi, and dt' = dt, we have d , ^ u' z = — (x — vt) = u, - v dt The transformations of all three components of velocity are as follows: u' x = u x — v u' y = u„ (6-6) «'s = "z Finally, diflferentiating these velocity components with respect to time, we have (for v = const.) three equalities involving the components of acceleration: du' x du x dUy dUy du' z du z dt' dt dt' dt dt' dt Thus the measure of any acceleration is the same in both frames: a' = a (6-7) Since this identity holds for any two inertial frames, whatever their relative velocity, we say that the acceleration is an invariant in classical mechanics. This result is the central feature of relativity in Newtonian dynamics (and it ceases to hold good in the description of motion according to special relativity). To illustrate the application of these ideas, consider the 175 The invariancc of Ncwton's law: relativity Fig. 6-7 Two dif- ferent views of the trajectory of an object after it has been released from rest with respect to a moving frame. S'. ide released from rest i n .V frame ■■■■ Frame S Frame S' BI simple and familiar example of a particle falling freely under gravity. Suppose that at t = 0, when the axes of the systems 5 and S' are coincident, an experimenter in S' drops a particle from rest in this frame. The trajectories of the particle, as seen in S and 5', are plotted in Fig. 6-7. In each frame the particle is observed to follow the expected trajectory according to the kinematic equations with a vertical acceleration g. In the S frame, the particle has an initial horizontal velocity and therefore it follows a parabolic path, whereas in S' the particle, under the aetion of gravity, falls straight down. Observers in these two different frames would agree that the equation F = ma, where they use the same F, accounts properly for the trajectories for any particle launehed in any manner in either frame. The frames are thus equivalent as far as dynamical experiments are con- cerned — either frame may be assumed stationary and the other frame in motion, with the same laws of mechanies providing correct explanations from the observed motions. This is a simple example of the invariance of Newton's law itself. INVARIANCE WITH SPECIFIC FORCE LAWS 1 In this seetion we shall consider a little more carefully what is involved in a transformation of Newton's second law of motion. What do we mean by a transformation of this law? Unless we have an explicit law of foree, F = ma can be regarded as only a preseription for deducing F from the observed motions. So let us consider the foree provided by the interaetion between two objects. Suppose, for simplicity, that the interaetion is provided by something like a stretehed rubber band, so that the foree is a funetion only of the distance r between the objects; i.e., we can put F = f (r). For further simplicity, let us assume that the 'This seetion can be omitted without loss of continuity. 176 Foree, inertia. and motion motion is confined to the x axis. Then the force exerted on object 2 by object I , as measured in frame S, can be written : F12 =/(*2- *i) (6-8) Newton's law, as applied to object 2, is then stated as follows in terms of measurements in S alone: F 12 = f(x? — x\) = /M2fl2 (6-9) We shall now rewrite this equation so that it is expressed entirely in terms of measurements made in the frame S'. From Eq. (6-5) we have X2 - Xi = (X2 + vl) - (x[ + ot) = x 2 - x\ Thus Fl2 = f(X2 ~ X[) but according to the assumed law of force, the function f(x' 2 — x[) is precisely the specification of the force F i2 in terms of measurements in S'. Hence we can put Fl2 = F\2 Turning now to the right-hand side of Eq. (6-9), the Galilean transformations give us a = a'; and in Newtonian dynamics the inertial mass is a constant: m 2 = m'i. Thus we are able to write F[ 2 = f(x 2 - x\) = m' 2 a 2 (6-10) We see here in explicit terms how the Newtonian law of motion is invariant with respect to the particular choice of inertial frame, provided that the Galilean transformations cor- rectly describe the transformations of displacements and times between one frame and another. A more complicated force law, as long as it involved only the relative positions and velocities of two interacting objects, would possess this same property of invariance. If, however, the force depended on absolute positions and velocities — e.g., if the force law were of the form F12 = f(xl - x?) then the form of the equation of motion would cease to be the same in all inertial frames. Nothing in our experience has re- vealed such a situation, which would make the laws of physics 177 Invariance with Specific force laws appear different in a laboratory and in a train or plane moving at constant velocity. If the physical laws were different for different observers, this might be a clue to the uniqueness of certain frames of reference. It was, indeed, believed for a long time that a unique reference frame must exist, in the form of a medium, pervading space, in which the waves of light could be carried. (Otherwise, how could light travel from the stars to the earth?) But Einstein showed how the equivalence of all inertial frames and the invariance of all physical laws could be pre- served, provided that the kinematics of Galileo and the dynamics of Newton were replaced by ncw formulations that merged into the old ones in the region of moderate or small velocities. NEWTON'S LAW AND TIME REVERSAL The subject of this section might be more dramatically, although less accurately, stated as a question: Is time reversible? Look at the two stroboscopic photographs in Fig. 6-8. The first shows an individual object moving vertically under gravity. Is it falling down or "falling up"? The second shows a collision between two objects. Which were the paths of the objects before collision? In both cases the answer must be "We don't know." A motion picture of these two sequences of events could be run backward and it would be impossible for the viewer to detect any violation of Newton's laws. The reason is that all velocities of a collection of particles can be reversed without violating Newton's law of motion. A time-reoersal operation (replacing t by — t in the kinematical equations) changes every u to -v, but it leaves the acceleration unchanged. (Gravity, for example, remains in the downward direction.) This is because acceleration involves the second derivative with respect to time. Thus any conclusions about forces that we reach as a result of watching a dynamical process in reverse sequcnce are identical with what we would conclude from the process itself. We do not see attractions apparently turn into repulsions, or anything like that. And yet, when we see an ordinary motion picture in reverse, it quickly becomes apparent from the behavior of inanimate objects — leaving aside the ludicrous effects of reversing human actions, which appear strange for quite different reasons — that most physical actions havc a well-defined direction. Imagine, for example, a sequence in which a glass falls from a table and shatters into small fragments on the floor. If wc saw a motion 178 Force, inertia, and motion t ■ V-/ l j - U (a) (b) Fig. 6-8 (a) Strohoscopic photograph ofan object moving vertically under gravily. Which way is i t moving — up, or down? (b) Slroboscopic photograph ofan elaslic collision. lis time sequence is lo all intents and purposes completely reversible. (From PSSC Physics, D. C. Heath, Lexinglon, Massachusells, 1965.) 179 Ncwton's law and time reversal picture in which the fragments gathered themselves together into a whole glass, which then jumped up onto the table, this would clearly be unbelievable— nature doesn't act like that. Yet a "micromovie" of the individual atomic encounters at every stage of the process ought to be perfectly time-reversible. Thus we are faced with a puzzle: Newton's law implies that the fundamental dynamical behavior of an individual particle is reversible in time, but when one takes a system of very large numbers of particlcs, apparently the behavior ceases to be time- reversible. The resolution of this mystery is found in the detailed statistical analysis of many-particle systems — the subject known as statistical mechanics. As long as we are dealing, as we shall be here, with systems of only a few particles, the problems associated with time rcvcrsal do not arise and we shall not con- sider them further. CONCLUDING REMARKS It will probably have become apparent to you during the course of this chapter that the foundation of classical mechanics, as represented by Newton's second law, is a complex and in many respects subtle matter. The precise content of the law is still a matter for debate, nearly three centuries after Newton stated the first version of it. In a fine diseussion entitled "The Origin " and Nature of Newton's Laws of Motion," one author (Brian Ellis) says: "But what of Newton's second law of motion? What is the logical status of this law? Is it a definition of foree? Of mass? Or is it an empirical proposition relating foree, mass, and acceleration?" 1 Ellis argucs that it is something of all of these: Consider how Newton's second law is actually used. In some fields it is unqueslionably truc that Newton's second law is used to define a seale of foree. How else, for example, can we mea- sure interplanetary gravitational forees? But it is also un- questionably true that Newton's second law is sometimes used to define a scale of mass. Consider, for example, the use of the mass spectrograph. And in yet other fields, where foree, mass, and acceleration are all easily and independently measurable, Newton's second law of motion funetions as an empirical cor- relation between these three quantities. Consider, for example, 'Published in a colleclion of essays, Beyond the Edge ofCeriainly (Robert G. Colodny, ed.), Prentice-Hall, Englewood Cliffs, N.J., 1965. 180 Foree, inertia, and motion PROBLEMS the application of Newton 's second law in ballistics and rock- etry . . . To suppose that Newton's second law of motion, or any law for that matter, must have a unique role that we can describe generally and call the logical status is an unfounded and unjustifiable supposition. Since forcc and mass are both abstract concepts and not objective realities, we might conceive of a description of nature in which we dispensed with both of them. But, as one physicist (D. H. Frisch) has remarked, "Whatever we think about ultimate reality it is convenient to follow Newton and split the description of our observations into 'forces,' which are what make masses accelerate, and 'masses,' which are what forces make accelerate. This would be just tautology were it not that the observed phe- nomena can best be classified as the result of different forces acting on the sameset of masses." Ellis spells out this same idea in more detail : Now there are, in fact, many and various procedures by which the magnitudes of the individual forces acting on a given system may be determined — electrostatic forces by charge and distance measurements, elastic forces by measurement of strain, magnetic forces by current and distance determinations, gravitational forces by mass and distance measurements, and so on. And it is an empirical fact that when all such foree measurements are made and the magnitude of the resultant foree determined, then the rate of change of momentum of the system under considera- tion is found to be proportional to the magnitude of this re- sultant foree. And so it is that we obtain an immensely fruitful and accurate description of a very large part of our whole experience of objects in motion, through the simple and compaet statement of Newton's second law. ' 6-1 Make a graphical analysis of the data represented by the strobo- scopic photograph of Fig. 6-2(a) to test whether this is indeed ac- celeration under a constant foree. J For further diseussion of these questions, the American Journal of Physics is a perennial source. See, for example, the following artieles: L. Eisenbud, "On the Ciassical Laws of Motion," Am. J. P/iys., 26, 144 (1958); N. Austern, "Presentation of Newtonian Mechanics," Am. J. Phys., 29, 617 (1961); R. Weinstock, "Laws of Ciassical Motion: What's F? What's ml What's a?", Am. J. Phys., 29, 698 (1961). 181 Problem-, 6-2 A cabin cruiser of mass 15 metric tons drifts in toward a dock at a speed of 0.3 m/scc after its engines have been cut. (A metric ton is 10 :i kg.) A man on the dock is able to touch the boat when it is 1 m from the dock, and thereafter he pushes on it with a force of 700 N to try to stop it. Can he bring the boat to rest before it touches the dock? 6-3 (a) A man of mass 80 kg jumps down to a concrete patio from a window ledge only 0.5 m above the ground. He neglects to bend his knees on landing, so that his motion is arrested in a distance of about 2 cm. With what average force does this jar his bone structure ? (b) If the man jumps from a ledge 1.5 m above the ground but bends his knees so that his centcr of gravity descends an additional distance h after his feet touch the ground, what must h be so that the average force exerted on him by the ground is only three times his normal weight ? 6-4 An object of mass 2 kg is acled upon by the following combina- tion of forces in the xy plane: 5 N at 6 = 0, 10 N at 6 = jt/4, and 20 N at 8 = 47r/3. The direction 6 = corresponds to the +x direc- tion. At t = the object is at the point j;=-6m and y = 3 m and has velocity components v x = 2 m/sec and v y = 4 m/sec. Find the object's velocity and position at t = 2 sec. 6-5 The graphs shown give information regarding the motion in the xy plane of three different particles. In diagrams (a) and (b) the small (a) Parabolic Note! ° (Vertical lines equally spaced) N Direction ,of motion (b) / = (c) 182 Force, inertia, and motion dots indicate the positions at equal intervals of time. For each case, write equations that describe the force components F x and F„. 6-6 An observer first measures the velocity of an approaching object to be 10~ 2 m/sec and then, 1 sec later, to be 2 X 10 -2 m/sec. No intermediate readings are possible because the observer's instruments take a full second to determine a velocity. If the object has a mass of 5 g, what conclusions can the observer make about (a) The size of the force that had been aeting? (b) The impulse supplicd by the force? (c) The work donc by the force? 6-7 A partiele of mass 2 kg oscillates along the x axis according to the equation -!) x = 0.2 sin 1 5/ where x is in meters and / in seconds. (a) What is the force aeting on the partiele at / = 0? (b) What is the maximum force that acts on the partiele? 6-8 A car of mass 10 3 kg is traveling at 28 m/sec (a little over 60 mph) along a horizontal straight road when the driver suddenly sees a fallen tree blocking the road 100 m ahead. The driver applies the brakes as soon as his reaetion time (0.75 sec) allows and comes to rest 9 m short of the trec. (a) Assuming constant deceleration caused by the brakes, what is the decelerating force? What fraetion is it of the weight of the car (take g = 9.8 m/sec 2 )? (b) If the car had been on a downward grade of sin -1 ^) with the brakes supplying the same decelerating force as before, with what speed would the car have hit the tree ? 6-9 A partiele of mass m follows a path in the xy plane that is de- seribed by the following equations: x = A(at — sin a*) y = A(l — cos at) (a) Sketch this path. (b) Find the time-dependent force veetor that causes this motion. Can you suggest a way of producing such a situation in praclice? 6-70 A piece of string of Iength /, which can support a maximum tension T, is used to whirl a partiele of mass m in a cireular path. What is the maximum speed with which the partiele may be whirled if the circle is (a) horizontal; (b) vertical? 183 Pro b lem s Part II Classical mechanics at work Does the engineer ever predict the acceleration of a given bodyfrom a knowledge of its mass and of the forces acting upon it ? Of course. Does the chemist ever measure the mass of an atom by measuring its acceleration in a given field offorce? Yes. Does the physicist ever determine the strength of a field by measuring the acceleration of a known mass in t ha t field? Certainly. Why then, should any one of these roles be singled out as the role of Newton's second law ofmotion? Thefact is that it has a variety of roles. brian ellis, The Origin and Nature of Newton's Laws of Motion (1961) Using Newton's law rr is worth reemphasizing the fact that Newton's law may be used in two primary ways: 1. Given a knowledge of all the forces acting on a body, we can calculate its motion. 2. Given a knowledge of the motion, we can infer what force or forces must be acting. ' This may seem like a very obvious and quite trivial separa- tion, but it is not. The first category represents a purely deductive activity — using known laws of force and making clearly defined predictions therefrom. The second category includes the in- duclive, exploratory use of mechanics — making use of observed motions to learn about hitherto unknown features of the inter- actions between objects. Skill in the deductive use of Newton's law is of course basi c to successful analytical and design work in physics and engineering and can bring great intellectual satisfaction. But, for the physicist, the real thrill comes from the inductive process of probing the forces of nature through the study of motions. It was in this way that Newton discovered the law of universal gravitation, that Rutherford discovered the atomic nucleus, and that the particle physicists explore the struc- ture of nucleons (although, to be sure, this last fleld requires analysis in terms of quantum mechanics rather than Newtonian mechanics). ■If the law of force is also known, this can be used to obtain information about an unknown mass. The quotation opposite treats this as a third category. 187 SOME INTRODUCTORY EXAMPLES Kinemalic = do + at equations : > s = so + eot + W 2 (Constant acceleration only) d 2 = do 2 + 2a(s — so) In Chapter 8 we shall have something to say about the way in which Newton arrived at his insight into gravitational forces from the study of planetary motions. But first we shall discuss how onc goes about calculating motions from given forces. There will be a certain lack of glamour about some of this initial work — but that is an inescapable aspect of science, as of every- thing else. And some systematic groundwork at the beginning will pay rich dividends later. We shall restrict ourselves at first to cases in which the forces are constant. Thus we shall be able to use the kinematic equations for constant acceleration [Eq. (3-10)]. For convenience, we quote the equations again here: (7-1) In any given problem our procedure will be first to identify all the forces acting on an object and, from F nct = ma, to cal- culate the resultant acceleration. We can then use the kinematic equations to describe the subsequent motion. Generally, this latter step is merely an exercise in mathematics; the real "physics" of the situation lies in the analysis of what forces are present. Do not be misled by the apparently trivial nature of these in- troductory problems. Taken at face value, they are, indeed, uninteresting and inconsequential. But the method of analysis is of salient importance— exactly the same approach is used in far more sophisticated problems. We purposely begin by choosing rather elementary systems to clarify the procedure, so the prob- lems are trite. But keep your eye on the method— it is powerful. Examp!e 1: Block on a smooth table. By a "smooth" surface we mean one that is incapable of exerting any force tangential to itself. No such surface exists but some come close enough for this to be a useful idealization. Consider a block of mass m pulled horizontally along a frictionless surface. Question: What is the motion of the block? ' Wc first must dctcrmine what forces act on the block. To assist the analysis, mentally "isolate the block" with an imag- ■In this and succeeding cxamplcs we make the assumption that an object at rest rclative to the earth's surface has zero acceleration. This is only approxi- mately true— see Chapter 12 for a full diseussion. 188 Usine Newton's law Fig. 7-1 (a) BIock pulled horizontally on a perfeclly smooth surface. (b) Same block with a slring puli in g in an arbitrary direction. inary boundary surface [see Fig. 7-1 (a)] and draw a sketch showing all the forces that act from outside through this surface on the block. 1 They are F„ = force of gravity T = tension in the string (a contact force) N = contact force the table exerts on the block; this force is normal to the surface, hence our choice of the letter N We have introduced vector symbols here because clearly forces acting in more than one direction are involved. The complete statement of Newton's law for this case is thus T + N + F„ = ma. Since we assume that the block has no acceleration in the vertical direction, we conclude that N and F„ are equal and opposite vectors; i n other words, N + F„ = Expressing this a diflferent way, we can say that the magnitude F of F„ is equal to the magnitude A' of N, because the mag- nitude of a vector, without regard to its direction, is defined as a positive quantity. The condition of zero net force component along y can then be stated in the equation N — F„ = 0. The sum of the forces on the block— the "net" force— is thus (by vector addition) equal to T, and hence we have T = ma (with both T and a horizontal). 'We shall, in these examples, treat each object as if it were a point particle, with all forces accordingly acting through a single point. 189 Some introductory cxamples I The problem becomes somewhat more substantial if we suppose that the puli of the string is not horizontal [Fig. 7-1 (b)]. The initial vector statement of Newton's law is of the samc form as before, but the analysis of it into vertical and horizontal components now gives us the following equations: Vertically : Horizontally : N + Tsind - F„ = Tcosf? = ma The first equation tells us the magnitude of N'. N = F g - Tsmd (The vertical component of the tension in the string hclps to support the block, and so A' becomes less than F g .) The second equation tells us directly the magnitude of the horizontal ac- ccleration. We notice that there is a physical limitation on this analysis. Unless the table can puli downward on the block, as well as being able to push upward on it to support its weight, the force N is necessarily upward, as shown in the diagram, and the scalar N is necessarily positive. Thus, if Tsin > F„, we cannot satisfy the assumed conditions. What happens then? Example 2: Block on a rough table. As in Example 1, we shall suppose that the block is pulled by a string with a force T, and we shall, for simplicity, take the puli to be horizontal (see Fig. 7-2). Although this is certainly not a difficult problem, we cannot say what will happen unless we have some information about the properties of the frictional force 3\ If it is "dry fric- tion," we can put 5 < fiN, where /* is the coefficient of friction. The value of A' is cqual to F a in this case, and so, writing F B as equal to m times the gravitational acceleration g, we have 9 < nmg. The inequality expresses the ability of the frictional force to adjust itself, up to a certain limit, to balance the force T. Fig. 7-2 Block pulled horizontally on a rough surface. 190 Using Newton's law Thus the equation T + JF = ma, which defines the horizontal acceleration, leads to two separate statements: If T < nmg: 7 - ff = If T > nmg: T — nmg = ma (and hence a = 0) [and hence a = (T/m) — fig] We shall not consider in detail what would happen if T were applied at an angle to the horizontal. But one can see that this will modify the value of N and hence, in turn, the limiting value of 5F. You should analyze this case for yourself. Example 3: Block on a smooth incline. Two forces act on the block (Fig. 7-3): the force N normal to the (frictionless) surface and the gravitational force F„. These are in different directions, they are bound to have a nonzero resultant, and the block must accelerate. We could, if we wished, introduce hori- zontal and vertical coordinates x and y and write equations for F = ma that would define the components a z and a y of the vector acceleration a : Ncosd — F„ = ma„ N siri 6 = ma x It is clear, however, that a is parallel to the slope, and it is simpler in this case to introduce the coordinate s representing distance along the slope in the downward direction. Resolving the forces along and perpendicular to 5, we have F„ sin 6 = ma N - F cos 6 = If we write the magnitude of the gravitational force as equal to m times the gravitational acceleration g, the first of these equa- tions gives us a = g sin d Fig. 7-3 Forces acting on a block on a perfectly smooth incline. 191 Some introductory examples Fig. 7-4 BIock at resi with respect to an accelerating elevator; the spring scale records its apparent weight. Example 4: BIock in an elevator. A block sits on a package scale (a spring scale!) in an elevator (Fig. 7-4). Question: What does the scale read as the elevator moves up and down? Mentally isolating the block, we recognize that only two forces act on it. They are: V a = the force of gravity (downward) N = the contact force (upward) exerted by the scale on the block The package scale records the magnitude of N, because, by the equality of action and reaction for objects in contact, the force exerted on the scale is —N. The value of this reading, under any conditions, is what we shall call the measwed weight {W) of the block. If the elevator (and the block and scale with it) has an acceleration a, measured as positive upward, then Newton's la w requires Thus, if the elevator is stationary or is moving with constant velocity upward or downward, the reading of the scale is equal to the gravitational force on the block. But if the elevator has a positive acceleration (upward), then we have N = m(g + a) In this case the weight of the block, as measured by the scale, is greater than the gravitational force on it. One will often 192 Using Newton's Iaw notice this effect personally, as an increased force on the soles of the feet, when riding an elevator in two situations: (a) when the elevator is picking up speed, going upward, and (b) when it is slowing down, going downward. Both of these involve upward acceleration . Similarly, if the elevator is slowing down in its upward motion or just beginning its downward motion, a is downward, so that the measured weight is less than F„. (Inci- dentally, the internal discomfort that one sometimes feels in an elevator can be linked directly to Newton's law. If the elevator acquires a positive (upward) acceleration, onc's heart and stomach must — literally — sink a little before they experience extra forces from the surrounding tissues to supply the accelera- tion called for. Example 5 : Two connected masses. Here is a simple example designed to illustrate the important point that one is free to isolate, in one's imagination, any part of a complete system, and apply F = ma to it alone. Figure 7-5 shows two masses con- nected by a light (massless) string on a smooth (frictionless) surface. A horizontal force, P, pulls at the right-hand mass. What can we deduce about the situation? First, we can imagine an isolation boundary drawn around both m, and m 2 and the string that connects them. The only external horizontal force applied to this system is P, and the total mass is rtii plus m 2 - Hence we have P = (mi + mi)a This at once tells us the acceleration that is common to both masses. Next, we can imagine an isolation boundary surrounding the connecting string alone. In Fig. 7-5 we indicate the forces T! and T 2 with which the string pulls on the masses; by the equality of aetion and reaetion, the string has forces equal to — Ti and — T 2 applied to its ends. The sum of these forces must equal the mass of the string times its acceleration a. Assuming Fig. 7-5 Two con- nected blocks pulled horizontal/y on a per- fectly smooth surface. Newton's law must apply to any part of the system that one chooses to consider. 193 Some introduetory examples the mass of the string to be negligible, this means that T] and T 2 would have the same magnitude, T, which we call the tension in the string. (This idealized result is, of course, rather obvious; what is worth noting is that, in any real situation, there would have to be enough differencc of tcnsions at the ends of the string to supply the requisitc accelerative force to the mass of the string itself.) Finally, we can imagine drawing isolation boundaries around m x and m 2 separately, and applying Newton's law to the hori- zontal motion of each: T = m\a P — T = mza Adding these equations, we arrive at the equation of motion of the total system once again. But if we take either equation alone, substitution of the already determined value of a will give the value of the tension T in terms of P and the masses. Once again we shall acknowledge the simple character of these problems. But if they are studied in the spirit in which they are offered— not for their own sake, but for the way in which they exemplify the systematic use of Newton's law — they will be found to suggest a sound approach to almost any problem that involves a direct application of F = mz. MOTION IN TWO DIMENSIONS Ali the examples in the last section dealt with motion along one dimension only. This is not a serious limitation because, thanks to the "independence of motions," we can always analyze a situation at any given instant in terms of the components of force and acccleration along separate coordinate directions. The simplest case of this occurs when the acceleration a is a constant vector. It is then very convenient to choose the direction of a itself as one coordinate; along any direction perpendicular to this the acceleration component is, by definition, zero, and the velocity component must be constant. Probably the most familiar example of this procedure is the analysis of motion under gravity in the idealization that air resistance can be ignorcd. We con- sidered this as a purely kinematic problem in Chapter 3. Another example, with the additional feature that we have control over 194 Using Newton's law (a) Accelerating anode Controlgrid- Focusing anode - Vertical deflectins plates Fluorescent screen Horizontal deflecting plates Glasswallof tube Vertical deflection plates (simplified) (b) Fluorescent screen Resultant deflection from central axis Fig. 7-6 (a) Diagram of the main features ofa simple cathode-ray tube. (b) Simplified diagram ofelectrical connections and electron trajectory. the applied force, is the motion of an electron beam in a cathode- ray tube. Let us consider this as a dynamical problem. ' Figure 7-6(a) shows a sketch of a cathode-ray tube, and Fig. 7-6(b) is a schematic diagram of some principal features. A well-focused beam from an electron gun passes between two pairs of deflection plates which, if appropriately charged, will deflect the beam transversely away from the central axis so as to strike any particular point on the fluorescent screen. We shall call the central axis the z direetion, so that the direetions of transverse deflection, perpendicular to z and to one another, can be called x and y, just as in a real oscilloscope. The electron gun accelerates the electrons through a po- 'We shall make use of certain results concerning the effects of electric forees. If these are not already familiar to you, see, for example, PSSC, Physics (2nd ed.), Part IV, Heath, Boston, 1965. 195 Motion in two dimensions tential difference K . This gives to each electron a kinetic energy eV (where e is the elementary charge) so that it acquires a lon- gitudinal velocity component v 2 given by j»n?, fi = eVo (7-2) Therefore, After leaving the gun, the electron is not subjected to further acceleration or decelcration along the z direction, so its z co- ordinate continues to change at the rate v z . We shall suppose that a potential difference V is applied between the upper and lower ^-deflection plates. This means that if an electron were to travel all the way from one plate to another, it would acquire an energy eV in consequence of the work done on it by the transverse electric force F y . If we suppose the plates to be parallel and spaced by a distance d that is small compared to their length, the force F v would have the same value at all points between the plates, so that the gain of energy would also be given by F v d. Hence we have F =^ * . d Given this value of F v the electron will, throughout its passage between the plates, have an upward acceleration given by * m md How far vertically will the electron be deflected? This will depend on the amount of time it spends between the deflection plates. If the horizontal extent of the plates is /, and the hori- zontal component of the electron's velocity has the constant value v z , the time is given simply by = ]_ Vz From Eq. (7-1) we cari determine the transverse displacement y that occurs during the time this upward force acts: s y = vo v t + \a y t y = + 2 md\vj 196 Using Newton's law But from Eq. (7-2), mv z 2 /e is just 2V , so we obtain , = IL (7-3) This expression gives the transverse displacement of the electron as it emerges from the deflection plates. The resultant displace- ment away from the central axis of the spot on the fluorescent screen may bc found from trigonometric considerations. The z and y components of the displacement of the electron while between the plates are given by z = o, t y = W 2 Eliminating t from these equations shows that y is proportional to z 2 , and the trajectory is therefore a parabola. Once the electron leaves the region between the plates, there is no further force on it and it travcls in a straight line at an angle equal to arctan (o„/o,), where v u is its transverse velocity component upon leaving the plates. Now we have eV i v y = aj = — ■% — v„ eVI VI v z mdo? IV ad The additional transverse deflection Y as the electron travels the distance D from the deflector plates to the screen is thus given by r-*?-g£ (7-4) v, 2Vod If D » /, as is usually the case in practice, most of the total transverse deflection is contained in Y, and we can use Eq. (7-4) to estimate the approximate sensitivity of an actual oscilloscope [sensitivity = (spot deflection)/(deflection voltage).] We may note in passing a striking feature of Eqs. (7-3) and (7-4). AU distinguishing characteristics of the particle as an electron have disappeared. Any negatively charged particle, with any charge or any mass, could in principle pass through the system and end up at the same spot on the screen. In a real oscilloscope, as Fig. 7-6(a) indicates, the deflector plates are not flat and parallel; thus Eqs. (7-3) and (7-4) do not strictly apply. It remains true, however, that these equations indicate the way in which the deflection depends on the ac- celerating and deflecting voltages, and also, in a less specific way, 197 Molion in two dimensions on the characteristic dimensions of the tube. MOTION IN A CIRCLE The problem of circular motion is often presented as though it were separate from the kinds of applications of Newton's law that we have discussed so far. It may therefore be worth em- phasizing that it involves a completely straightforward use of F = wa. The only special feature is that the radial component of the acceleration is uniquely related to the radius of the path and the instantaneous speed. Consider first the simplest case, in which we suppose that an object of mass m is traveling around a circle of radius r at a constant speed v (Fig. 7-7). Such a motion can be set up, for example, by tethering a puck, by means of a taut string, to a fixed peg on a very smooth horizontal surface (e.g., an air table) and giving the puck an arbitrary velocity at right angles to the string. Then, as wc saw in Chapter 3, the acceleration of the object is purely toward the center of the circle and is of mag- nitude o' 2 /r. We know that the production of this "centripetal" acceleration necessitates the existencc of a corresponding force: F = ma = (7-5) In the hypothetical case that we have described, this force would have to be supplied by the tension T in the tethering string. And if the string is not strong enough to supply a force of the required magnitude, it will break and the object, being now free of forces (at least in the plane of the motion), will fly off along a tangent. Fig. 7-7 Basic dynamical situa- tion for a parlicle traveling in a circle at constant speed. 198 Using Newton's law \ y n* \ \ S* > j To center of curve \ 1 u «e / ~^j / / / / v y /r F. (b) Rjf. 7-5 (a) View ofa curve in a road, as seenfrom vertically ocerhead. (b) Car on the banked curve, as seen front directly behind or m front. The motion ofa car on a banked curve presents an important practical problem of this type. Suppose that a road has a radius of curvature r (measured in a horizontal plane), as indicated in Fig. 7-8(a) and is banked at an angle a as shown in Fig. 7-8(b). A car, traveling into or out of the plane of the latter diagram with speed v, has a centripetal acceleration v 2 /r. The purpose of banking is to make it possible for the car, traveling at some reasonable speed, to be held in this curved path by a force exerted on it purely normal to the road surface — i.e., thcre would be no tangential force, as there would have to be if the road surface were horizontal. (And the inability of the road-tire contact to supply such a tangential force would rcsult in skidding.) Consider, then, the ideal case as shown in Fig. 7-8(b). Resolving the forces vertically and horizontally, and applying F = wa, we have Nsina = r (7-6) Ncosa - F„ = Replacing F a by mg, and solving for a, we find that tana = — gr (7-7) which defines the correct angle of banking for given values of v 199 Motion in a circlc and r. Altematively, given r and a, Eq. (7-7) defines the speed at which the curve should be taken. The situations that arise if greater or lesser values of v are used will require the introduction of a frictional force J perpendicular to N (and of limiting mag- nitude ij.N) acting inward or outward along the slope of the banked surface (see Problem 7-17). CURVILINEAR MOTION WITH CHANGING SPEED If a particle changes its speed as it travels along a circular path, it has, in addition to the centripetal acceleration toward the center of the curvature, a component of acceleration tangent to the path. This tangential acceleration component represents the rate of change of the magnitude of the vclocity vector. (This is in contrast to the centripetal acceleration component, which de- pends upon the rate of change of the direction of the velocity vector.) We derivcd the relevant results in Chapter 3 [Eq. (3-18)]. The situation is most easily handled by considering the two components separately: radial component of acceleration (at right J = a, = - angles to the path) 1 j (r is the radius of r curvature) transverse acceleration) ,- £o do [note that this is } = a$ = lim — = — « (tangent to the path)| a<-.o Al di the change of magnitude (only) of the velocity vector] (7-8) With only slight rcinterpretation, we may apply these results to the case of motion along any arbitrary curvilinear path. For every point along such a path, therc is a center of curvature and an associatcd radius of curvature (both of which change as one movcs along the path). Provided that we interpret the symbols r and u to mean the instantaneous values of the radius of curvature and the speed, the above expressions are perfectly general. They give the instantaneous acceleration components — tangent to the ' In our formal diseussion of coordinate systems in Chapter 2 we defined the outward radial direction as positive. We are introducing here the symbol a, to denote the acceleration component along this direction. The minus sign indicates that this acceleration is in faet toward the center (i.e., centripetal), as we have diseussed. Sce also the diseussion on pp. 106-108. 200 Using Nevvton's law Arbitrary path Fig. 7-9 Acceleration componenls at a poinl on an arbitrarily curved palh. path, and normal to the path — for the general curvilinear motion. In this case Fig. 7-9 indicates a more appropriate notation for these acceleration components. A particle of dust that rides, without slipping, on a phono- graph turntable as it starts up provides a simple and familiar example of a particle possessing both radial and tangential ac- celeration components. Its total acceleration a (Fig. 7-10) is the combination of a r and o» at right angles: a = (a r 2 + a» 2 ) 1 ' 2 . The net horizontal force applied to the particle by the turntable must be in the direction of a and of magnitude ma. If the contact between the turntable and the particle cannot supply a force of this magnitude, such circular motion is not possible and the particle will slip relative to the surface. If we analyze the process of whirling an object at the end of a string, so as to bring it from rest up to some high speed of circular motion (as, for example, in the athletic event "throwing the hammer") we see that the string must perform two functions: (1) supply the tangential force to increase the speed of the object, and (2) supply the force of the appropriate magnitude mv 2 /r Fig. 7-10 Net acceleration vector ofa particle attached to a disk ifthe angular velocity of the disk is changing. 201 Curvilinear motion with changing speed Fig. 7-11 Pari ide tracels i n a circular path about O. To in- crease its speed, the slring PO' must provide a force com- ponent langenlial to the circle. toward the center of the circle. To fiil this dual role, the puli supplied by the string must "lead" the object, so that the tension supplied by it has a tangential component, as shown in Fig. 7-11. This can be maintained if the inner end, O', of the string is con- tinually moved around in a circular path, as indicated. This is the kind of thing we do more or less instinctively in practice. CIRCULAR PATHS OF CHARGED PARTICLES IN UNIFORM MAGNETIC FIELDS One of the most important examples of circular motion is the behavior of electrically charged particles in magnetic fields. The separate section on pp. 205-206 summarizes the properties of the magnetic force and describes how this force is in general given by the following equation: F mag = W X B (7-9) where q is the electric charge of the particle (positive or negative). Let us now imagine a charged particle of (positive) charge q, moving in the plane of this page, in a region in which the magnetic fieid B points perpendicularly down into the paper. Then from Eq. (7-9) we have F = qvB (7-10) This is a pure deflccting force, always at right angles to the particle's motion. Hence the magnitude of the velocity v cannot change, but its direction changes uniformly with time. Thus, although there is no center of force in the usual sense, the par- ticle describes a circular path of radius r with center O [Fig. 202 Using Newton's law (a) (b) Fig. 7-12 (a) Charged parlicle following a circular path in a uniform magnelicfield. (b) Magnelicalty curved tracks ofelectrons and posilrons in a cloud chamber. The main fealure is the pai h ofan eleclron of inilial energy 30 MeV going through a succession ofalmosl circular orbils ofdecreasing radius as U loses energy. (Courtesy oflhe Unicersily of California Lawrence Radiation Laboralory, Berkeley.) 7-1 2(a)]. The centripetal acceleration is v 2 /r, and so by Newton's law we have mu qvB = whence mo (7-11) Thus the radius of the circle is a mcasure of the momentum mv 203 Charged particles in uniform magnetic ftelds of a particle of given charge in a given magnetic field. This fact underlies the nuclear physicist's method for determining the momenta of charged particles in a cloud or bubble chamber [see Fig. 7-1 2(b)]. From the radius of the circular track which a charged particle generates in a cloud chamber placed between the poles of a magnet the momentum can be readily found if the charge of the particle is known. To get the period T (or the angular velocity), wc write qvB = mwv or o> = % = 1 B (7-12) which is independent of the particle's spced. The angular fre- quency given by Eq. (7-12) is called the "cyclotron" frequency and depends, for a fixed magnetic field, only on the ratio of charge to mass of the particle. It was the recognition of this by E. O. Lawrence that led him in 1929 to design the first cyclotron, in which protons could be raiscd to high energies by the application of an alternating electric field of constant frequency. The hold- ing of charged particles in an orbit of a given radius by means of a magnetic field is an essential feature of most high-energy nuclear accelerators. Having the velocity in a plane perpendicular to B is, of course, a very special case. But if this condition is not satisfied, we can imagine v to be resolved into one component perpendic- ular to B and a second component parallel to B. The latter, by Eq. (7-9), has no magnetic forcc associated with it; the former Fig. 7-13 Helical path of charged par- ticle having a velocity component parallel t o a magnetic field. v s = const. 204 (Jsing Newton's lavv changes with time in precisely the way described above. Thus the resultant motion is a helix, whose projection on the plane perpendicular to B is a fixed circle, as shown in Fig. 7-13. CHARGED PARTICLE IN A MAGNETIC FIELD It is a matter of experimental fact that an electrically charged particle may, at a givcn point in space, experience a force when it is moving which is abscnt if it is at the same point but sta- tionary. The cxistence of such a force depends on the presence somewhere in the neighborhood (although perhaps quite far away) of magnets or electric currents. Detailed observations reveal the following featurcs [cf. Fig. 7-14]: 1 . The force is always exerted at right angles to the direction of the velocity v of the particle. The force is reversed if the direction of the velocity is reversed. 2. The force is proportional to the amount of charge, q, carried by the particle. The force reverses if the sign of the charge is reversed. 3. For a given value of q and a given direction of v, the size of the force is proportional to the magnitude of v. 4. For motion parallel to one certain direction at a given point, the force is zero. This direction coincides with the direc- tion in which a compass needle placed there would align itself. It is called the direction of the magnetic _/?e W at that point. 5. For any other direction of motion, the direction of the Fig. 7-14 (o) Situa- tions of zero force and maximum force for a charged par- ticle moving in a magnet ic field. (b) General veetor relationship of veloc- ity, magnetic field, and magnetic force. Field direction Field direction (b) 205 Charged particle in a magnetic field force is perpendicular to the plane formed by v and the field direction. 6. The magnitude of the force is proportional, for given values of q and v and for a fixed magnetic field arrangement, to the sine of the angle between v and the field direction. Ali the above results can be summarized in a very compaet mathematical statement. We are going to intro'duce a quantitative measure of the magnetic field strength and denote it by the veetor symbol B, in such a way that Fig. 7-14 shows the relation of the direetions of qv, B, and F for a charged partiele. (Note that q may be of either sign and that F is normal to the plane containing v and B.) Then the value of F, in both magnitude and direction, is given by the following equation, involving the cross produet of veetors (see p. 127): F = «vXB We can then use this equation to define the quantitative measure of the magnetic field strength, such that a field of unit strength, applied at right angles (sin 6 = 1) to a charge of 1 C moving with a speed of 1 m/sec, would produce a force of 1 N. MASS SPECTROGRAPHS The charaeteristie curvature of a partiele of given q/m in a magnetic field provides the basis of nearly all methods of obtain- ing precise relative values of atomic and isotopic masses. Such devices often make use of the magnetic force in two ways, first as a velocity selector and second in the manner deseribed in the last seetion. The velocity selection takes advantage of the faet that the magnetic force on a charged partiele is proportional to v, whercas the electric force is not. If a beam of charged partieles travels between parallel plates a distance d apart, with voltage difference V, the electric force is given by F a = ^ (7-13) But the magnetic force is given by Eq. (7-10), fmag = $<-'£ These forees can be arrangcd to be in opposite direetions by 206 Using Newton's law Fig. 7-15 (a) Schematic design ofa simpleform ofmass spectrometer. (b) Exampfe ofisolopic separalion and mass analysisfor isolopes of xenon with a spectrometer like that in (a). [After A. O. Nier, Phys. Rev., 52, 933 (.1937).] 207 Mass spcctrographs applying the magnetic field at right angles to the electric field. There is then only one speed at which particles can travel through these "crossed fields" undeflected: V_ Bd (7-14) Figure 7-15(a) is a diagram of a simple mass spectrograph that uses a velocity filter of this type, followed by a region in which the chargcd particles (ions) travel in a semicircular path under the influence of the magnetic field alone. Figure 7— 15(b) shows an example of the separation of isotopes by such a device. THE FRACTURE OF RAPIDLY ROTATING OBJECTS The question of the stresses set up in a rotating object, and the possibility of fracture i f they bccome excessive, provides another good example of Newton's law applied to uniform circular motion. Whenever an object such as a wheel is rotating, every portion of it has an acceleration toward the axis of rotation, and a corresponding accelcrative force is required. Suppose, for example, that a thin wheel or hoop of radius r is rotating about its axis at n revolutions per second (rps) [see Fig. 7-1 6 (a)]. Then any small section of the hoop, such as the one shown shaded, must be supplied with a forcc cqual to its mass, m, multiplied by its centripetal acceleration v 2 /r. In this case the magnitude of the angular velocity w is defined by the equation to = 2irn Fig. 7-16 (a) Rotat- ing hoop. (i) Forces aeting on a small ele- men! of the hoop. 208 Using Newton's law Thus the instantaneous specd is given by v = ar = lirrn (7-15) If we make an isolation diagram [Fig. 7-16(b)] for the small portion of material shown shaded in Fig. 7-16(a), it is clear that the forces acting on it must be supplied via its contact with adjacent material of the rim. ' These forces must, by symmetry, be tangential to the rim at each point. (For example, if the force exerted on Am at one end had a component radially outward, then by the equality of action and reaction the portion of ma- terial with which it was in contact would be subjected to a force with a component radially inward. But in a uniform hoop, all portions such as Am are equivalent; there is no basis for any asymmetries of this kind.) Thus we can picture the small portion of the rim being acted on by a force of magnitude T at each end. If the length of arc represented by this portion is As, it subtends an angle A0, equal to As/r, at the center O, and each force has a component equal to Tsin(A0/2) along the bisector of AS. Thus by Newton's law we have o 2 27"sin(A0/2) = Am — Putting sin(A0/2) « A0/2, this then gives us 2 7"A0 ^ Am — r i.e., T — ~ Am — r r or 7" A.? = o 2 Am (7-16) Let us now express the mass Am in terms of the density, p, of the material and the volume of the piece of the rim. If the cross-sectional area o f the rim is A, we have Am = pAAs 'We shall assume that the spokcs of the wheel serve primarily to give just a geometrical connection between the rim and the axis and have almost negli- gible strength. 209 The fraeture of rapidly rotating objects Substituting this into Eq. (7-16) and substituting o = lirrn from Eq. (7-15), we obtain the following result: T = 4t¥rV< CM7) Now, it is an cxperimental fact that a bar or rod of a given material will fracture under tension if the ratio of the applied force to the cross-sectional area exceeds a certain critical value — the ultimate tensile strength, S. Thus we can infer from Eq. (7-17) that a hoop of the kind we have bcen discussing has a critical ratc of rotation, above which it will burst. We have, in fact, Wmox = iw 1/2 (7-18) Suppose, for cxample, that we have a steel hoop of radius 1 ft (i.e., about 0.3 m). The density of steel is about 7600 kg/m 3 , and its ultimate strength is about 10° N/m 2 . These values lead to a value of n max of about 500 rps, or 30,000 rpm— much faster than any such wheel would normally be driven. However, the rotors of ultracentrifuges arc, in fact, driven at speeds of this order, up to a significant fraetion of the bursting speed for their particular radius (sec p. 513). MOTION AGAINST RESISTIVE FORCES We shall now consider an important class of problems in which an object is subjected to a constant driving force, F , but has its motion opposed by a resistive force, R, that always acts in a direetion opposite to the instantaneous velocity. Typical of such forees are the frictional resistance as an object is pulled along a solid surface, or the air resistance to falling raindrops, moving cars, and so on. In general this resistive force is a funetion of speed, so that the statement of Newton's law must be written as follows: F - R(o) = m j (7-19) As we saw in Chaptcr 5, the resistive force of dry frietion is in fact nearly independent of u, as indicated in Fig. 7-1 7(a), so that we can put R(c) -= 5 ~ const. 210 Usili»; Newton's la w Fig. 7-17 (o) Dric- mg and resistive forces for an object resisted by dryfric- lion. (b) Dricing and resistive forces for an object resisted by fluid friction. Resistive force RW Driving force F„ Driving force / ff Frictional force "m (a) (b) ■■■■■MHHHnfflBH MBBHHBHBimiHHi Equation (7-19) then reduces to the simplc case of acceleration undcr a constant net force, as dcscribed by the kinematics of Eqs. (7-1). The situation is very different in the case of fluid resistance, for which R(o) increases monotonically with o, as indicated in Fig. 7-1 7(b) and as described [Eq. (5-4)] by the relation R(c) = Ac + Bv 2 (7-20) In this case, if we consider an object starting out from rest under the force F , the initial acceleration is FJm, but the net driving force is immediately rcduced to a value below F , because as soon as the object has any appreciable velocity it is exposed, in its own frame of reference, to a wind or flow of fluid past it at the speed v. The statement of Newton's law as it applies to this problem must now be written m— = Fq — Av — Bu dt (7-21) (In this equation F , A, B, and v are all taken to be positive. One must be careful to consider what is the appropriate state- ment of Newton's law for this system if the direetion of v is reversed.) The solving of Eq. (7-21) is not at all such a simple matter as our familiar problems involving constant forces or forces perpendicular to the velocity. Wc are now faced with finding the solution to an awkward differential equation. We do not intend to plunge into all the formal mathematies of this problem. In- stead, this is a suitable moment to point to the value of ap- 211 Motion against resistive forces proximate numerical methods— in other words, the method of the digital computer, using finitely small intervals. We outlined the principle of the method in Chapter 3 [Eq. (3-11)]; here we have a good case for using it. First, however, let us consider some individual features of the solution: 1. For some rangc of small values of v, the acceleration will be almost constant and o will start off as a linear funetion of t with slope F /m. 2. As c increases, a decreases monotonically, giving a steadily decreasing slope in the graph of v versus t (Fig. 7-18). 3. There is a limiling speed, v m , under any given applied foree. It is the speed at which the graph of R(v) versus v is inter- seeted by a horizontal line at the ordinate equal to F [see Fig. 7-17(b)]. Algebraically, it is the positive root of the quadratic equation B v* Ao - F = (What is the status of the negative root, and why is it to be discarded?) Notice the contrast between the sharply defined value of v„, in the graph of R(v) versus o, and the gradual manner in which this velocity is approached (and in principle never quite reached) if one considers u as a funetion of time (Fig. 7-18). It is well worth taking a moment to consider the dynamical situation represented by v = v,„. It is a motion with zero ac- celeration under zero net foree, but it seems a far cry from the unacceleratcd motion of objects moving under no foree at all; and it is ccrtainly not an application of what we understand by the principle of inertia. But let us emphasize that, like static equilibrium, it is a case of ZF = 0. Every time we see a car Fig. 7-18 Asymplolic approach lo terminal speed for object in a fluid resislice medium. 212 Using Newton's law hurtling along a straight road at a steady 80 mph, or a jet plane racing through the air at a constant 600 mph, we are seeing objects traveling under zero net force. This basic dynamical fact tends to be obscured because what matters in practical terms is the large value of the driving force F needed to maintain the steady motion once it has been established. DETAILED ANALYSIS OF RESISTED MOTION In order to see how Newton's law for resisted motion [Eq. (7-21)] works in practice we need to introduce one additional feature already describcd in Chapter 5 — that the two terms in the rc- sistive force differ not only in their dependcncc on d but also in their dependence on the linear dimensions of the object (see Fig. 7-19). Specifically, for a sphere of radius r we have A = C\r B = C 2 r' 2 and thus R(v) = C\ro + C2'V (7-22) The two terms in the resistance thus become equal at a critical speed, v c , defined by the formula V c = c 2 r (7-23) We know that the term proportional to v will dominate the picture if v is sufficiently small (since Bv 2 /Av ~ v), but we know Fig. 7-19 (a) Linear and quadratic terms influid resistance for a small object, with viscous resistance predominant. (b) Similar graphfor a large object, for which the v 2 term predominates at a/l except low speeds. 213 Dciailcd analvsis of resisted motion equally that for speeds much in excess of v c (say v > 10u c ) the quadratic term takes over. If the resistive medium is air, the coeffkients c t and c 2 have the following approximate values: ci = 3.1 X 10- 4 kgm- 1 sec _1 c 2 = 0.87 kg/m 3 Thus, if r is expressed in meters, we have ... 3.6 X 10~ 4 ,, ... v c (m/sec) = (7-24) This means that for an object such as a small pebble, with r » 1 cm, the value of v c is only a few centimeters per second; a speed equal to 10 times this (say 0.5 m/sec) would be acquired in free fail under gravity within a time of about 0.05 sec and a distance of about 0.5 in. (Check thcsc numbers!) Thus for most problems of practical interest (we shall consider an exception in the next section) the motion under a constant applied force in a resisting medium can be extremely well described with the help of the following simplified version of Eq. (7-21): m ^ « F - B 2 (7-25) di The resistive term, Bv 2 , is quite important in the motion of ordinary objects falling through the air under gravity. This becomes very apparent if we calculate the terminal speed, v t , by putting dv/dt = in Eq. (7-25), with F set equal to the gravita- tional force, mg. Take, for example, the case of our pebble of radius 1 cm. The density of stone or rock is about 2.5 times that of water, i.e., about 2500 kg/m :! , so we have m = ^ pr :s « 4(2.5 X 10 a ) X (10" 2 ) 3 « 10" 2 kg The value of F is thus about 0.1 N. The value of the coef- ficient B (= c 2 r 2 ) for an object of this size is about 10 -4 kg/m. Substituting these values in the equation F - Bu, 2 = we find Vi « 30 m/sec Under the assumptions of genuinely free fail, this speed would 214 Using Ncwton's law be attained in a time of about 3 sec and a vertical distance of about 150 ft. We can be sure, then, that the effects of the re- sistance become quite significant within times and distances appreciably Iess than these. This means that many of the ideal- ized problems of free fail under gravity, of the sort that everyone meets in his first encounter with mechanics, do not correspond very well with reality. Let us now consider the computer procedure for solving these problems. Given a computer, it is almost as little trouble to handle the full equation [Eq. (7-21)] as it is to use the ap- proximation represented by Eq. (7-25). But for the purposes of discussing the method, we shall use the simpler, approximate form. We choose some convenient small increment of time, A/, and will measure time from t = in terms of integral multiples (n) of Af. In the simplest approach, we assume that the accelera- tion during each small interval remains constant at the value calculated for the beginning of that interval. Thus, for an initial velocity equal to zero, the acceleration during the first At is set equal to a (= F J m). The velocity Vi at the end of this interval is thus given by ci = + aoA/ Using this velocity, we calculate the acceleration d\ at / = At: B 2 a i = tfo Cl m Applying this acceleration to the next interval, the velocity at t = 2 At is given by 02 = vi + fli At We know that the first step in the calculation, as performed in this way, leads to an overestimate of Pj, but we see that in this particular problem the error is in some measure compensated by leading in turn to an underestimate of a%. The calculated values of a in successive time intervals are as indicated in Fig. 7-20(a). Applying an exactly similar treatment to the changes of position (jt„) we would take our set of values of velocities, as given by the above calculation and represented graphically in Fig. 7-21 (a), and would use the following formula : x„ +i = x„ + v„At 215 Detailed analysis of resisted motion Fig. 7-20 (d) Basis ofsimple ealculation of acceleration versus time for objecl start- ingfrom resi i n a resistive medium. The acceleration is calculaied from the speed al the begin- ning ofeach time in- terval. {b)lmproved approach to the same problem, using ac- celeration calculated from speed at mid- point ofeach time interval. Thus we should have x\ = ro A/ = (clearly an underestimate) X2 = x\ + oi At = vi At X3 = X2 + CiAt Our graph of x versus t would be the result of summing the areas of the rectangles in Fig. 7-21 (a) up to successively greater values of n. A more sophisticated analysis takes account of the fact that a better average value of the acceleration or velocity during a given time interval is provided by the instantaneous value at the midpoint of that interval. Thus the acceleration between n At and (n + 1) At is set equal to the instantaneous value at (n + -J) At. This Ieads to the following formulas: Fig. 7-21 (a) Veloc- ity-time graph based on velocities at the beginnings of the suc- cessive time intervals. (b) Improved graph based on velocities evaluated at mid- points of the time intervals. 216 Using Newton's law v n +i = c„ + a n+ i /2 A/ *»+l = X„ + Vn+i /2 &l (7-26) This looks fine, but we now run into a slight snag when we try to get the calculation started. To find v u the velocity at the cnd of the first interval, we now need v (= 0) and a ll2 . The lattcr, however, by Eq. (7-25), depends on a knowledge of v U2 , which we do not yet know. (Noticc that in the first, crude method, we were able to start out directly from the initial conditions v and a .) We are thus forced to compromise a little, although we Fig. 7-22 (a) Com- parison ofidealized (resistanceless) and aclual dependence of speed on lime for a falling pebble of radius l cm. (b) Idealized and aclual dislances fallen I by such a pebble. 217 Detailcd analysis of resistcd motion are still Ieft with a better treatment than before. What we do is to calculate an approximate value of vn 2 from the equation V112 » «o -z (7-27) and then we are under way. Figures 7-20(b) and 7-21 (b) show what this method means in terms of graphs of a and v against time. Figure 7-22 shows graphs of the actual calculated variation of speed and distance with time for our 1-cm-radius pebble falling in air, under gravity; the idcalizcd frce-fall curves are given for comparison. MOTION GOVERNED BY VISCOSITY If we are dealing with microscopic or near-microscopic objects, such as particlcs of dust, then, in contrast to the situations dis- cussed above, the resistance is due almost entirely to the viscous term, Av, up to quite high values of v. If, for cxample, we con- sider a tiny particle of radius 1 m (= 10 -6 m), then Eq. (7-24) tells us that the critical spccd, v c , at which the contributions Av and Bo 2 bccome equal, is 360 m/sec. This implies a wide range of lower speeds for which the motion is controlled by viscous resistance alone, and the statcmcnt of Newton's law can be writtcn, without any appreciable error, in the following form: m *° mFo -Ao (7-28) di with A = c\r Motion under these conditions played the central role in R. A. Millikan's cclebrated "oil-drop" experiment to determine the elcmentary electric charge. The basic idea was to measure the electric foree exerted on a small charged object by finding the terminal spced of the object in air. If the radius of the particle is known, the resistive foree is completely determined, and the driving foree must be equal and opposite to it at v = v t . In Millikan's original experiments the charged partieles were tiny droplets of oil in the mist of vapor from an "atomizer." Such droplets havc a high probability of carrying a net electric 218 Using Newton's law o d q Fig. 7-23 Basic paratlel-plate arrangement in Millikan experiment. charge of either sign when they are produced. In order to apply electric forces to them, Millikan used the arrangement shown schematically in Fig. 7-23. Two parallel metal plates, spaced by a small fraction of their diameter, are connected to the terminals of a battery. ' The force on a particle of charge q anywhere be- tween these plates is given by _ v (7-29) where V is the voltage difference applied to the plates and d is their separation in meters. If q is measured in coulombs, F is given in newtons by this equation. (See also our discussion of the cathode-ray oscilloscope, p. 195.) Thus if a dynamic balance were set up between this electric driving force and the resistive force Av-, we should have V q-j= Ad, = ciru, a The droplets randomly produced in a mist of oil vapor are of various sizes. The ones that Millikan found most suitable for his experiments were the smallest (partially because they had the lowest terminal speed under their own weight). But these drop- lets were so tiny that even through a medium-power microscope they appeared against a dark background merely as points of light; no direct measurement could be made of their size. Millikan, however, used the clever trick of exploiting the law of viscous resistance a second time by applying it to the fail of a droplet under the gravitational force alone, with no voltage between the plates. Under these conditions we have Fo = mg = y pr g where p is the density of oil (about 800 kg/m 3 ). The terminal speed of fail under gravity is then given by ■Millikan himself used plates about 20 cm across and 1 cm apart and several thousand volts. Most modern versions of the apparatus use smaller values of all three quantities. 219 Motion governed by viscosity 4jt a y pr g = cirD B (Gravitational) u„ = y — r 2 (7-30) Putting in the approximate numcrical values we find (Gravitational) i?„ ~ 10 r 2 (u ( in m/sec, /• in m) Putting r « lju = 10 -0 m, we have f « 10 -4 m/sec = 0.1 mm/sec Such a droplet would take over 1 min to fail 1 cm in air under its own weight, thus allowing prccision measurements of its speed. (It is clear, incidentally, that for such motions as this the re- sistive term Bo 2 is utterly negligible.) It is worth noting the dynamical stability of this system, and indeed of any situation involving a constant driving force and a resistive force that increases monotonically with speed. If by chance the droplet should slow down a little, there is a net force that will speed it up. Conversely, if it should speed up, it is subjected to a net retarding force. If one could observe the motion of a falling droplet in sufficient detail, this behavior would doubtless be found, because the air, being made up of individual molecules, does not behave as a perfectly homogeneous fluid. In other words, the speed of the droplet would fluetuate about some average value, although the fluetuations would be exceedingly small. In the Millikan experiment proper, the vertical motion of the charged droplets is studied with the electric force either aiding or opposing the gravitational force. Thus if we measure velocities as positive downward, the terminal velocity in both magnitude and sign will be defined by the equation mg + *r- cm, (7-31) where V is the voltage of the upper plate relative to the lower and q is the net charge on the drop (positive or negative). Al- though in principle the terminal velocity is approached but never quite reached (see Fig. 7-18), the small droplets under the con- ditions of the Millikan experimcnt do, in effect, reach this speed within a very short time — much less than 1 msec in most cases (see the next seetion). Millikan was able to follow the motion of a given droplet 220 Using Ncwton's lavv for many hours on end, using its electric charge as a handle by which to puli it up or down at will. In the course of such pro- tracted observations the charge on the drop would often change spontaneously, and several different values of v, would be ob- tained. The crucial observation was that in any such experiment, with a given value of the voltage V, the speed v t was limited to a set of sharp and distinct values, implying that the electric charge itself comes in discrete units. But Millikan went further and obtained the first precise value of the absolute magnitude of the elementary charge. ' You might wonder why Millikan used such a roundabout method to measure the electric force exerted on a charged particle. After all, it would in principle be possible to hang such an electrified particle on a balance and measure the force in a static arrangement. However, in practice, when only a few elementary charges are involved, the forces are extremely weak and such a method is not feasible. For example, the force on a particle with a net charge of 10 elementary units, between plates 1 cm apart with 500 V between them, is only about 10~ 13 N, equal to the gravitational force on only 10 ,upg! GROWTH AND DECAY OF RESISTED MOTION We have seen how the velocity under a constant force in a resistive fluid medium rises asymptotically toward the terminal value. What happens if the driving force is suddenly removed? We can guess that the velocity will decay away toward zero in a similar asymptotic way, as indicated in Fig. 7-24. If the initial speed is small enough, the whole decay depends on viscous re- sistance alone and is governed by a special, simplified form of Eq. (7-28) in which F is set equal to zero: m— = — Av dt or (7-32) do — = — at) dt where a = A/m. 'For his own full and interesting account (and much other good physics), see R. A. Millikan, The Electron (J. W. M. Du Mond, ed.), University of Chicago Press (Phoenix Series), Chicago, 1963. 221 Growth and decay of rcsisted motion Fig. 7-24 Growth and decay of the velocity ofa parlicle controlled by a vis- cous resistiue force proportional to v. You may recognizc Eq. (7-32) as the basic differential equation of all forms of exponential decay. Whether you do or not, you may like to see how this equation can be solved by what is in effect just arithmetic, using the approach of numerical analysis and digital-computer techniques. We divide up the time into a large number of equal intervals Al, and interpret Eq. (7-32) as telling us that the changc of v between t = n At and t = (n + 1)A/ is proportional to the mean velocity during that interval: Au = v n+ i -*~-"\—2—) At Solving this, we have t;„ + i 1 - aAt/2 _ 1 + a Al/2 where/is a constant ratio, less than unity. The velocities at equal intervals of time thus decrease in geometric proportion. If the initial velocity is u , the velocity at time t (= k Al) is given by Substituting the expression for / from the preceding equation and putting k = t/At, we thus get o(t) fO 1 - aAr/2 \ 1 + a At/l) l/Ai We shall now consider what happens to this expression as we imagine the intervals At to be made shorter and shorter, and their number correspondingly greater. To simplify the discussion, we shall put aAt = - z 222 Using Newton's law The quantity z is then a large number that we shall allow to approach infinity. Besides substituting oi/= l/z in the above equation for u(l), we shall also replace the exponent t/At by its equivalent quantity, azt. Thus we have /l - \/2z\ a " \TTtj2z) \i -r i/^z/ i.e., o (0 «» tfO/> where Let us look at the behavior of p(z) as z is made larger and larger (Table 7-1). As z increases, the number p is clearly ap- TABLE 7-1 z y(z) Decimal value of p 1 2 3 4 5 0/3)' (3/5) 2 (5/7)3 (7/9)* (9/1 l) 5 0.3333 . . . 0.3600 0.3644 0.3659 0.3667 10 (19/21)'» 0.3676 proaching a limiting value; this value is 0.367879 . . . and is the reciprocal of the famous number e (= 2.71828 . . .), which forms the base of natural or Napierian logarithms. Thus in Eq. (7-32) we can put lim p(z) = 0.367879 . . . = e -1 Z— *X and hence we have the following expression, now exact, for the value of v(i): v(t) = v e~°' (7-34) The reciprocal of a in Eq. (7-34) is of the dimension of time 223 Growth and decay of resisteci molion and represents a characteristic time constant, t, for the ex- ponential decay pjocess. I n the particular case of small spheres moving through the air, r is defined by the equation _ m m_ 0. A ar We can express this in much more vivid terms by introducing the terminal velocity of fail under gravity, v g . For from Eq. (7-30) we have mg "'w It follows that we have r = 2» • (7-35) g Thus t is equal to the time that a particle would take to reach a velocity equal to the terminal Velocity under conditions of free fail. For an oil drop of radius 1 /j., with v„ » 0.1 mm/sec, this time would be only about 10 -5 sec (= 10/tsec). In a time equal to any substantial multiple of t, the value of v(t) as given by Eq. (7-34) falls to a very small fraction of vo- For example, if we take the basic equation v(t) = voe-' ir . (7-36) and put t = 10r, then the value of v becomes less than 10 -4 of v , which for many purposes can be taken to be effectively zero. It is more or less intuitively clear that the growth of velocity toward its limiting value, after a particle starts from rest under the action of a suddenly applied driving force, must be a kind of upside-down version of the decay curve (see Fig. 7-24). In fact, if the terminal velocity is o«, the approach to it, under these conditions of viscous rcsistance, is described by v(l) = o Al - e-" T ) (7-37) If one chooses to write this in the form ■ t v — lir v, — u(0 - »( e one can see explicitly how closely the growth and decay curves are related. Indeed, one could almost deduce Eq. (7-37) from Eq. (7-36) plus the recognition of this symmetry. 224 Using "New lon's law AIR RESISTANCE AND "INDEPENDENCE OF MOTIONS" When Galileo put forward the proposition that the motion of a projectile could be analyzed into separate horizontal and vertical * parts, with a constant velocity horizontally and a constant ac- celeration under gravity vertically, he made a great contribution to the conceptual basis of mechanics. It may be worth pointing out, however, that this "independence of motions" breaks down if one takes into account the resistive force exerted on objects of ordinary size. We have seen how for such objects the magnitude of the resistive force is proportional to the square of the speed. Consider an object moving in a vertical plane (Fig. 7-25). At a given instant let its velocity v be directed at an angle 6 above the horizontal, as shown." The object is then subjected to the gravitational force, F„, and a resistive force R(u), of magnitude Bv 2 , in a direction opposite to v. Newton's law applied to the x and y components of the motion at any instant, thus gives us m d -^= -R z = -Bo 2 cose at ■ dCy _ 2 • n m-^ = —mg — Bu sin 6 Since v cos 6 = v z , and v sin 6 = v u , we can rewrite these equa- tions as folio ws: do * la \ m — = - {BojOm m-— = -mg - (flOPy dt Thus the equation governing each separate component of the velocity involves a knowledge of the magnitude of the total velocity and hence of what is happening at each instant along the other coordinate direction. The larger the magnitude of v, the more important does this cross connection between the dif- ( a ) ferent components of the motion become. Thus we really cannot caleulate the vertical motion of a falling body without reference Fig. 7-25 Resistive and gravitational force vectorsfor a partiele moving in a vertical plane. 225 Air resistance and "independence of motions" to the horizontal component. Recognition of this fact may be salutary for those who are accustomed to taking for granted the idealized equations of falling objects. In the folklore of physics there is the story of the impecunious student who would make money off his nonscientific acquaintances by a bet. He asked them to consider two identical objects. One is dropped from rest at a certain height above the ground; the other is fired off horizontally at the same instant. Which one will reach the ground first? The victim (poor ignoramus!) would often have some vague idea that the fast horizontal motion of the second object must somehow keep it in the air longer. But of course that is demonstrably false, isn't it?— look at the analysis of motion under gravity in Chapter 3. And even a direct test would appear to prove the poin t if the velocities were kept small. But the complete equations, as written above, show that a large initial horizontal velocity would increase the time taken to descend a given vertical distance. Moral: Beware of facile idealizations. Notice that if the resistance is purely viscous, varying as the first power of v, then the x and y components of the motion can be handled entirely separately, even though the problem is not that of idealized free fail. Thus, although one can always take a statement of Newton's law as the starting point of any problem of a particle exposed to forces, the way of proceeding from there to the analysis of the complete motion may depend critically on the precise nature of the individual problem. SIMPLE HARMONIC MOTION One of the most important of all dynamical problems is that of a mass attracted toward a given point by a force proportional to its distance from that point. If the motion is assumed to be confined to the x axis we have F(x) = -kx ( 7 ~ 38 ) A good approximation to this situation is provided by an object on a very smooth horizontal table (e.g., with air suspension) and a horizontal coiled spring, as shown in Fig. 7-26(a). The object would normally rest at a position in which the spring is neither compressed nor extended. The force brought into play by a slight displacement of the mass in either direction is then well described by Eq. (7-38). The constant k is called the spring 226 Using Newton's law (a) o Fig. 7-26 (a) Mass-spring sysiem on a frictionless horizontal surface. (b) Mass hanging front a vertical spring. (c) Graph of force versus displacement with a linear force law; the equilibrium silualions O and O' correspond to (a) and (b), respeclivety. constant and is measured in newtons per meter. This linear force law for a spring was discovered by Robert Hooke in 1676 and is named after him. ' An even simpler arrangement in practice, although slightly more complicated theoretically, is to suspend a mass at the bottom end of a vertical spring, as in Fig. 7-26(b). In this case the normal resting situation already involves an extension of the spring, sufficient to support the weight of the object. A further displacement up or down from this equilibrium position leads, however, to a net restoring force exactly of the form of Eq. (7-38). This is indicated in Fig. 7-26(c), which shows the magnitude of the force exerted by the spring as a funetion of its extension y. If one takes as a new origin of this graph the point O', then one has a net restoring force (F — mg) propor- tional to the extra displacement (y — ya). The great importance of this dynamical problem of a mass on a spring is that the behavior of very many physical systems under small displacements from equilibrium obeys the same basic equation as Eq. (7-38). We shall diseuss this in more detail in Chapter 10; for the present we shall just concern ourselves with solving the problem as such. J He first announced it in a famous anagram— ceiiinosssttuv— which 2 years later he revealed as a Latin sentence, "ut tensio, sic vis" ("as. the cxtension, so the force"). By this device (popular in his day) Hooke could claim prior publication for his discovery without actually telling his competitors what it was! 227 Simple harmonic motion This makes a good case in which to begin with the computer method of solution rather than with formal mathematics. The basic equation of motion, expressed in the form ma = F, is as follows: d\ . (7-39) Rewriting this as d 2 x k dfi m X we recognize that k/m is a constant, of dimension (time) 2 . Denoting this by w 2 , our basic equation thus becomes dfi = —W X (7-40) We can read this as a direct statement of the way in which dx/dt is changing with time and can proceed to calculate the ap- proximate change of dx/dt in a small interval of time At : i( i) - - di \<//^ and so (f)~- w! 2 CO X xM (7-41) This is the reverse of the process by which d 2 x/dt 2 was originally defined (cf. Chapter 3). Fig. 7-27 Displace- ment versus time in simple harmonic motion. 228 Using Newton's law Suppose, to be specific, that we start out at f = with x = x and v (= dx/di) = v both positive, as shown in Fig. 7-27. Then at time / = A/ we have x *» xo + f o Ar rfx 2 » — « i>o - w A;oAr The displacement is a little bigger, the slope a little less. Using these new values, we take another step of A/, and so on. Several features can be read from Eq. (7-41): 1. As long as x is positive, the slope dx/dt decreases when we go from t to / + A/. That is, \idx/dt is positive it gets smaller; if dx/dt is negative it becomes more negative. 2. The rate of change of dx/dt is proportional to x. The graph has its greatest curvature at the largest x, and as x — ► it becomes almost a straight line. 3. As soon as x becomes negative, dx/dt becomes less negative or more positive with each time increment A/. Using these considerations, we can construct the picture of a curve that is always curving toward the line x = (i.e., the t axis), necessarily forming a repetitive wavy pattern. Now anyone who has ever drawn graphs of trigonometric functions will recognize that Fig. 7-27 looks remarkably like a sine or cosine curve. More specifically, it suggests a comparison with the following analytic expression for the distance x as a function of time : x = A sin(af + ^o) where A is the maximum value attained by x during the motion, a a constant with the dimension (time) -1 , and <p an adjustable angle that allows us to fit the value of x at t = 0. Testing this trial function against the original differential equation of motion [Eq. (7-40)] requires differentiating x twice with respect to /: v = — = aA cos(atf + <^o) at .2 a = -— = —oc A sin(a/ + <f>o) — — a x at 2 We see that the solution does indeed fit, provided that we put a = w. This then brings us to the following final result: 229 Simple harmonic motion „v 1/2 x(t ) = A sin(w/ + *?o) where w = I — J (7-42a) -ay Equation (7-42a) is the characteristic equation of what is called simple harmonic motion (SHM), and any system that obeys this equation of motion is called a harmonic oscillator. The constant A is the amplitude of the motion, and v>o (Gk: phi) is what is called the initial phase (at / = 0). The complete argument, at + ip , of the sine function is called just "the phase" of the motion at any given instant. The result represented by Eq. (7-42a) could be equally well expressed by writing x as a cosine function, rather than a sine function, of /: x(l) = A cos(at + <p' ) (7-42b) with an appropriate value of the constant <p' . This form of the solution is found more convenient for some purposes. The harmonic motion is characterized by its period, T, which defines successive equal intervals of time at the end of which the state of the motion reproduces itself exactly in both dis- placement and velocity. The value of T is readily obtained from Eqs. (7-42) by noting that each time the phase angle (at + ip ) changes by 2w, both x and v have passed through a complete cycle of variation. Thus we can put <pi = uti + </>o tpi + 2vr = o>(/i + T) + vo Therefore, by subtraction, 2w = 10T or /' (r) ' (7-43) !-(!)' The form of this result corresponds to one's commonsense knowledge that if m gets bigger the oscillation goes more slowly, and if the spring is madc stiffcr (largcr k) the oscillation becomes more rapid. Example. A spring of negligible mass hangs vertically from a fixed point. When a mass of 2 kg is hung from the bottom end of the spring, the spring is stretched by 3 cm. What is the period of simple harmonic vibration of this system? First we calculatc the foree constant k. The gravitational 230 Using Newton's law force on the mass is 2 X 9.8 N = 19.6 N. This force causes an extension of 0.03 m; therefore, k = 19.6/0.03 = 653 N/m. Using Eq. (7-43) we then have . 1/2 "" sec '-*(&) =°- 35! (You might like to note that we need not have specified the mass. Any spring that extends 3 cm when a certain mass is hung on it will have a vibration period of 0.35 sec with this same mass. Why?) MORE ABOUT SIMPLE HARMONIC MOTION Fitting the initial conditions We have come to recognize, both as a general principle and through various examples, that the complete solution of any problem in the use of Newton's law requires not only a knowledge of the force law but also the speciflcation of two independent quantities that correspond to the constants of integration intro- duced as we go from a (= d 2 x/dt 2 ) to x. Most commonly we have talked of giving the initial position, x , and the initial velocity, v . Here, in our analysis of the motion of a harmonic oscillator, we also need initial conditions or their equivalent. Actually they appear as the two constants A and <p >n Eq- (7-42). We have already identified A as the amplitude of the motion and <p as the initial phase. If we are given the values of x and v we can readily solve for A and <po as follows: From Eq. (7-42a), x = A sin(co/ + ifio) = ^ = wA cos(o>/ + *>o) (7-44) dt Therefore, xo = A sin v>o Vq = o>A cos v>o It follows that 1/2 '-k+(?)T (7-45) oixo tan^o = oo 231 More about simple harmonic motion A geometrical representation of SHM There is a very basic connection between simple harmonic motion and uniform motion in a circular path. This fact leads to a particularly simple way of displaying and visualizing SHM. Imagine that a horizontal disk, of radius A, rotates with constant angular speed u about a vertical axis through its center. Suppose that a peg P is mounted on the rim of the disk as shown in Fig. 7-28(a). Then, if the disk is viewed edge on (horizontally) the peg will seem to move back and forth along a horizontal straight line [Fig. 7-28(b)]. Its motion along this line will correspond exactly to Eq. (7-42a) if the angular position of the peg at t = is correctly chosen so that the angle SOP in Fig. 7-28(c) is equal to ut + <pq. In this representation the quantity w is seen as the actual angular velocity of P as it travels around the circle. The peg has a velocity v that changes direction but always has the magnitude u A. Resolving v parallel to Ox at once gives Eq. (7-44). Thus the motion of the point B in Fig. 7-28(c) corre- Fig. 7-28 (a) Peg on a uniformly rotating disk. (b) Displace- ment of the peg as viewed in a plane con- laining the disk. (c) Detailed relation of circular motion to simple harmonic motion. 232 Using Newton's !aw sponds in every respect to that of a particle performing SHM along the x axis. Dynamical relation between SHM and circular motion Although the analysis just described is a purely geometrical one, it suggests a close dynamical connection, also, between the actual linear motion of a harmonic oscillator and the projection of a uniform circular motion. The acceptance of F = ma means that the same motion implies the same force causing it, whatever the particular origin of the force may be. This equivalence can be understood with the help of Fig. 7-29. A particle of mass m is kept moving in a circle of radius A by means of a string attached to a fixed support at the center of the circle, O. If the particle has a constant speed, v, the tension, T, in the string must be given by T = mv ~A Here v 2 / 'A is the magnitude of the instantaneous acceleration, a„ of m toward the center of the circle. Now the total force and the total acceleration at any instant can be resolved into their x and y components in a rectangular coordinate system. (Normally we would not be interested in doing this, because T and a, have well-defined constant mag- Fig. 7-29 Dynamical relationship between circular motion and linear harmonic motion. 233 More about simple harmonic motion PROBLEMS nitudes.) Taking the x components alone, we have 2 mu F x = - T cos e = — cos e A 2 v a x = r cos 8 A Thus the x component of the complete vector equation, F = ma, is F x = ma x , with the values of F x and a x stated above. In order to display the dynamical identity of this component motion with SHM, we can take the expressions for F x and a x separately, introducing the angular velocity « and putting v = uA. We then have F x - —mco 2 Acosd = —mw 2 x a x = —u 2 Acos8 = — u 2 x The first of these equations defines a restoring force proportional to displacement, exactly in accord with our initial statement of Hooke's law [Eq. (7-38)]. The second corresponds exactly to the equation [Eq. (7-40)] that was our starting point for the kinematic analysis of the problem: d x 2 Thus we see that the dynamical correspondence is complete in every respect. It tells us, moreover, that we could, if we wished, go the other way and treat a uniform circular motion as a super- position of two simple harmonic motions at right angles. This is, in fact, an extremely important and useful procedure in some contexts, although we shall not take time to follow it up here and now. 7-1 Two identical gliders, each of mass m, are being towed through the air in tandem, as shown. Initially they are traveling at a constant speed and the tension in the tow rope A is 7"o. The tow plane then begins to accelerate with an acceleration a. What are the tensions in A and B immediately aftcr this acceleration begins? 234 Using Newton's law M 7-2 Two blocks, of masses M = 3 kg and m = 2 kg, are in contact on a horizontal table. A constant horizontal force F = 5 N is applied to block M as shown. There is a constant frictional force of 2 N be- tween the table and the block m but no frictional force between the table and the first block M . (a) Calculate the acceleration of the two blocks. (b) Calculate the force of contact between the blocks. 7-3 A sled of mass m is pulled by a force of magnitude P at angle 8 to the horizontal (see the figure). The sled slides over a horizontal surface of snow. It experiences a tangential resistive force equal to M times the perpcndicular force W exerted on the sled by the snow. (a) Draw an isolation diagram showing all the forces exerted on the sled. (b) Write the equations corresponding to F = ma for the horizontal and vertical components of the motion. (c) Obtain an expression for the horizontal acceleration in terms of P, 6, m, n, and g. (d) For a given magnitude of P, find what value of 6 gives the biggest acceleration. 7-4 A block of mass mi rests on a frictionless horizontal surface; it is connected by a massless string, passing through a frictionless eyelet, to a second block of mass mi that rests on a frictionless incline (see the figure). (a) Draw isolation diagrams for the masses and write the equa- tion of motion for each one separately. (b) Find the tension in the string and the acceleration of mi. (c) Verify that, for 6 = jt/2, your answers reducc to the ex- pected results. 7-5 In the figure, P is a pulley of negligible mass. An external force F acts on it as indicated. (a) Find the relation between the tensions on the right-hand and left-hand sides of the pulley. Find also the relation between F and the tensions. (b) What relation among the motions of m. M, and P is pro- vided by the prescncc of the string? (c) Use the above results and Newton's law as applied to each block to find the accelerations of m, M, and P in terms of m. M, g, and F. Check that the results make sense for various specialized or simplified situations. 7-6 A man is raising himsclf and the platform on which he stands with a uniform acceleration of 5 m/sec 2 by means of the rope-and- pulley arrangement shown. The man has mass 100 kg and the plat- form is 50 kg. Assumc that the pulley and rope are massless and move without frietion, and neglect any tilting effects of the platform. 235 Problems Assume g = 10m/sec 2 . (a) What are the tensions in the ropes A, B, and C? (b) Draw isolation diagrams for the man and the platform and draw a separate force diagram for each, showing all the forces acting on them. Label each force and clearly indicate its direction. (c) What is the force of contact exerted on the man by the platform ? 7-7 In an equal-arm arrangement, a mass 5/no is balanced by the masses 3mo and 2mo, which are connected by a string over a pulley of negligible mass and prevented from moving by the string A (see the figure). Analyze what happens if the string A is suddenly severed, e.g., by means of a lighted match. "S - 5»i„ 3»i„ 2m 7-8 A prisoner in jail decides to escape by sliding to freedom down a rope provided by an accomplice. He attaches the top end of the rope to a hook outside his window; the bottom end of the rope hangs clear of the ground. The rope has a mass of 10 kg, and the prisoner has a mass of 70 kg. The hook can stand a puli of 600 N without giving way. If the prisoner's window is 15 m above the ground, what is the least velocity with which he can reach the ground, starting from rest at the top end of the rope? 7-9 (a) Suppose that a uniform rope of length L, resting on a fric- tionless horizontal surface, is accelerated along the direction of its length by means of a force F pulling it at one end. Derive an expression for the tension T in the rope as a function of position along its length. How is the expression for Tchanged if the rope is accelerated vertically in a constant gravitational field ? (b) A mass M is accelerated by the rope in part (a). Assuming the mass of the rope to be w, calculate the tension for the horizontal and vertical cases. 7-10 In 1931 F. Kirchner performed an experiment to determine the charge-to-mass ratio, e/m, for electrons. An electron gun (see the figure) produced a beam of electrons that passed through two "gates," each gate consisting of a pair of parallel plates with the upper plates connected to an alternating voltage source. Electrons could pass 236 Using Newton's law -07 -K o- / straight through a gate only if the voltage on the upper plate were momentarily zerc. Wilh the gates separated by a distance / equal to 50.35 cm, and with a gate voltage varying sinusoidally at a frequency / equal to 2.449 X 10 7 Hz (1 Hz = 1 cycle/sec), Kirchner found that electrons could pass completely undeflected through both gates when the initial accelerating voltage (K ) was set at 1735 V. Under these conditions the flight-time between the gates corresponded to one half-cycle of the alternating voltage. (a) What was the electron speed, deduced directly from / and /? (b) What value of e I m is implied by the data ? (c) Were corrections due to special relativity significant? [For Kirchner's original paper, see Ann. Physik, 8, 975 (1931).] 7-11 A certain loaded car is known to have its center of gravity half- way between the front and rear axles. It is found that the drive wheels (at the rear) start slipping when the car is driven up a 20° incline. How far back must the load (weighing a quarter the weight of the empty car) be shifted for the car to get up a 25° slope? (The distance between the axles is 10 ft.) 7-12 A child sleds down a snowy hillside, starting from rest. The hill has a 15° slope, with a long stretch of level field at the foot. The child starts 50 ft up the slope and continues for 100 ft on the level field before coming to a complete stop. Find the coefficient of friction between the sled and the snow, assuming that it is constant throughout the ride. Neglect air resistance. 7-13 A beam of electrons from an electron gun passes between two parallel plates, 3 mm apart and 2 cm long. After leaving the plates the electrons travel to form a spot on a fluorescent screen 25 cm farther on. It is desired to get the spot to deflect vertically through 3 cm when 100 V are applied to the deflector plates. What must be the accelerating voltage Vo on the electron gun? (Show first, in general, that if the linear displacement caused by the deflector plates can be neglected, the required voltage is given by V o = V(lD/2Yd), where Y is the linear displacement of the spot on the screen. The notation is that used on p. 197.) 237 Problem s 7-14 A ball of mass m is attached to one end of a string of length /. It is known that the string will break if pulled with a force equal to nine times the weight of the ball. The ball, supported by a frictionless table, is made to travel a horizontal circular path, the other end of the string being attached to a fixed point O. What is the largest number of revolutions per unit time that the mass can make without breaking the string? 7-75 A mass of 100 g is attached to one end of a very light rigid rod 20 cm long. The other end of the rod is attached to the shaft of a motor so that the rod and the mass are caused to rotate in a vertical circle with a constant angular velocity of 7 rad/sec. (a) Draw a force diagram showing all the forces acting on the mass for an arbitrary angle of the rod to the downward vertical. (b) What are the magnitude and the direction of the force exerted by the rod on the mass when'the rod points in a horizontal direction, i.e., at0 = 90°? 7-16 You are fiying along in your Sopwith Camel at 60 mph and 2000-ft altitude in the vicinity of Saint Michel when suddenly you notice that the Red Baron is just 300 ft behind you fiying at 90 mph. Recalling from captured medical data that the Red Baron can with- stand only 4 g's of acceleration before blacking out, whereas you can withstand 5 g's, you decide on the following plan. You dive straight down at full power, then level out by fiying in a circular are that comes out horizontally just above the ground. Assume that your speed is constant after you start to puli out and that the acceleration you experience in the are is 5 g's. Since you know that the Red Baron will follow you, you are assured he will black out and crash. Assuming that both planes dive with 2 g's ucceleration from the same initial point (but with initial speeds given above), to what altitude must you deseend so that the Red Baron, in trying to follow your subsequent are, must either crash or black out? Assuming that the Red Baron is a poor shot and must get within 100 ft of your plane to shoot you down, will your plan succeed ? After starting down you recall reading that the wings of your plane will fail off if you excced 300 mph. Is your plan sound in view of this limitation on your plane? 7-17 A curve of 300 m radius on a level road is banked for a speed of 25 m/sec (« 55 mph) so that the force exerted on a car by the road is normal to the surface at this speed. (a) What is the angle of bank ? (b) The frietion between tires and road can provide a maximum tangential force equal to 0.4 of the force normal to the road surface. What is the highest speed at which the car can takc this curve without skidding? 7-18 A large mass M hangs (stationary) at the end of a string that 238 Using Ncwton's law passes through a smooth tube to a small mass m that whirls around in a circular path of radius / sin 6, where / is the length of the string from m to the top end of the tube (see the figure). Write down the dynamical equations that apply to each mass and show that m must complete one orbit in a time of 2ir(lm/gM) v2 . Consider whether there is any restriction on the value of the angle 6 in this motion. 7-19 A model rocket rests on a frictionless horizontal surface and is joined by a string of length / to a fixed point so that the rocket moves in a horizontal circular path of radius /. The string will break if its tension exceeds a value T. The rocket engine provides a thrust F of constant magnitude along the rocket's direction of motion. The rocket has a mass m that does not decrease appreciably with time. (a) Starting from rcst at / = 0, at what later time / 1 is the rocket traveling so fast that the string breaks ? Ignore air resistance. (b) What was the magnitude of the rocket's instantaneous net acceleration at time 1 1/2? Obtain the answer in terms of F, T, and m. (c) What distance does the rocket travel between the time t\ when the string breaks and the time 2/i? The rocket engine continues to operate after the string breaks. 7-20 It has been suggested that the biggest nuclear accelerator we are likely to make will be an evacuated pipe running around the earth's equator. The strength of the earth's magnetic field at the equator is about 0.3 G or 3 X 10 -5 MKS units (N-sec/C-m). With what speed would an atom of lead (at. wt. 207), singly ionized (i.e., carrying one elementary charge), have to move around such an orbit so that the magnetic force provided the correct centripetal acceleration? (e = 1.6 X 10~ 10 C.) Through what voltage would a singly ionized lead atom have to be accelerated to give it this correct orbital speed ? 7-21 A trick cyclist rides his bike around a "wali of death" in the form of a vertical cylinder (see the figure). The maximum frictional force parallel to the surface of the cylinder is equal to a fraction n of the normal force exerted on the bike by the wali. (a) At what speed must the cyclist go to avoid slipping down? 239 Problems (b) At what angle (<p) to the horizontal must he be inclined? (c) If m ~ 0.6 (typical of rubber tires on dry roads) and the radius of the cylinder is 5 m, at what minimum speed must the cyclist ride, and what angle does he make with the horizontal? 7-22 The following expression gives the resistive force exerted on a sphere of radius r moving at speed v through air. It is valid over a very wide range of speeds. R(c) = 3.1 X 10 -4 r«J + 0.87/-V where R is in N, r in m, and v in m/sec. Consider water drops falling under their own weight and reaching a terminal speed. (a) For what range of values of small r is the terminal speed determined within 1% by the first term alone in the expression for R(v)l (b) For what range of values of Iarger r is the terminal speed determined within 1% by the second term alone? (c) Calculate the terminal speed of a raindrop of radius 2 mm. If there were no air resistance, from what height would it fail from rest before reaching this speed? 7-23 An experiment is performed with the Millikan oil-drop ap- paratus. The plates are 8 mm apart. The expcriment is done with oil droplets of density 896 kg/m 3 . The droplets are timed between two horizontal lines that are 2.58 mm apart. With the plates un- charged, a droplet is observed to take 23.6 sec to fail from one line to the other. When the upper plate is made 1100 V positive with respect to the lower, the droplet rises and takes 22.0 sec to cover the same distance. Assume that the resistive force is 3.1 X 10~ 4 n; (MKS units). (a) What is the radius of the droplet? (b) What is its net charge, measured as a number of elementary charges? (e = 1.6 X 10" 19 C.) (c) What voltage would hold the droplet stationary? [Use the precise value of the charge deduccd in part (b)]. 7-24 Two solid plastic spheres of the same material but of different radii, R and 2R, are used in a Millikan expcriment. The spheres carry equal charges q. The Iarger sphere is observed to reach terminal speeds as follows: (1) plates uncharged: terminal speed = vo (down- ward), and (2) plates charged: terminal speed = lm (upward). As- suming that the resistive force on a sphere of radius r at speed o is ciro, find, in terms of i>o and vi, the corresponding terminal speeds for the smaller sphere. 7-25 Analyze in retrospect the legcndary Galilean experiment that took place at the leaning tower of Pisa. Imagine such an experiment done with two iron spheres, of radii 2 and 10 cm, respectively, dropped simultaneously from a height of about 15 m. Make caleulations to 240 Using Ncwion's lavv r rn: n _L 2 m (a) determine, approximately, the difference in the times at which they hit the ground. Do you think this could be detected without special mcasuringdevices? (Density of iron « 7500 kg/m 3 .) 7-26 Estimate the terminal speed of fail (in air) of an air-filled toy balloon, with a diameter of 30 cm and a mass (not counting the air inside) of about 0.5 g. About how long would it take for the balloon to come to within a few percent of this terminal speed? Try making some real observations of balloons inflated to different sizes. 7-27 A spring that obeys Hooke's law in both extension and com- pression is extended by 10 cm when a mass of 2 kg is hung from it. (a) What is the spring constant k ? (b) The spring and the 2-kg mass are placed on a smooth table. The mass is pulled so as to extend the spring by 5 cm and is then released at / = 0. What is the equation of the ensuing motion? (c) I f, instead of being released from rest, the mass were started off at x = 5 cm with a speed of 1 m/sec in the direetion of increasing x, what would be the equation of motion? 7-28 When the mass is doubled in diagram (a), the end of the spring deseends an additional distance //. What is the frequency of oscillation for the arrangement in diagram (b)? AH individual springs shown are identical. 7-29 Any object, parlially or wholly submerged in a fluid, is buoyed up by a foree equal to the weight of the displaced fluid. A uniform cylinder of density p and length / is floating with its axis vertical, in a fluid of density po. What is the frequency of small-amplitude vertical oscillations of the cylinder? 7-30 (a) A small bead of mass m is attached to the midpoint of a string (itself of negligible mass). The string is of length L and is under constant tension T. Find the frequency of the SHM that the mass deseribes when given a slight transverse displacement. (b) Find the frequency in the case where the mass is attached at a distance D from one end instead of the midpoint. 7-31 A block rests on a tabletop that is undergoing simple harmonic motion of amplitude A and period T. (a) If the oscillation is vertical, what is the maximum value of A that will allow the block to remain always in contaet with the table? (b) If the oscillatior. is horizontal, and the coefficient of frietion betwcen block and tabletop is p., what is the maximum value of A that will allow the block to remain on the surface without slipping? 7-32 The springs of a car of mass 1 200 kg give it a period when empty of 0.5 sec for small vertical oscillations. (a) How far does the car sink down when a driver and three passengers, each of mass 75 kg, get into the car? 241 Problems (b) The car with its passengers is traveling along a horizontal road whcn it suddenly runs onto a piece of new road surface, raised 2 in. above the old surface. Assume that this suddenly raises the wheels and the bottom ends of the springs through 2 in. before the body of the car begins to movc upward. In the ensuing rebound, are the passengers thrown clear of their seats? Consider the maximum acceleration of the resulting simple harmonic motion. 242 Using Newton's law If it universally appears, by experiments and astronomical obseruations, that all bodies about the earth gravitate towards the earth . . . in proportion to the auantity of matter that they severally contain; that the moon likewise . . . gravitates towards the earth . . . and all the planets one towards another; and the comets in like manner towards the sun; we must, in conseauence of this rule, universally allow that all bodies whatsoever are endowed with a principle of mutual gravitation. newton, Principia (1686) 8 Universal gravitation THE DISCOVERY OF UNIVERSAL GRAVITATION in chapters 6 and 7 we have built up the kind of foundation in dynamics that Newton himself was the first to establish. In a nutshell, it is the quantitative identification of force as the cause of acceleration, coupled with the purely kinematic problem of relating accelerations to velocities and displacements. We shall now consider, as a topic i n its own right, the first and most splendid example of how a law of force was deduced from the study of motions. It is convcnient, and historically not unrcasonable, to con- sider separately three aspects of this great discovery: 1. The analysis o f the data concerning the orbits of the planets around the sun, to the approximation that these orbits are circular with the sun at the center. Several people besides Newton were closely associated with this problem. 2. The proof that gravitation is universal, in the sense that the law of force that governs the motion of objects near the earth's surface is also the law that controls the motion of celestial bodies. It seems clcar that Newton was the true discoverer of this result, through his analysis of the motion of the moon. 3. The proof that the true planetary orbits, which are ellipses rather than circles, are explained by an inverse-square law of force. This achievement, certainly, was the product of Newton's genius alone. In the present chapter we shall be able to discuss the first of these questions quitc fully, using only our basic results in the 245 kinematics and dynamics of particles. The second question requires us to learn (as Newton himself originally had to) how to analyze the gravitating properties of a body, like the earth, which is so obviously not a geometrical point when viewed from close to its surface. We shall present one approach to the prob- lem here and complcte the story in Chapter 11, where this special feature of the gravitational problem is discussed. The third question, concerning the exact mathematical description of the orbits, is something that we shall not go into at all at this stage; such orbit problems will be the exclusive concern of Chapter 13. THE ORBITS OF THE PLANETS We have described in Chapter 2 how the knowledge of the motions of the classical planets — Mercury, Venus, Mars, Jupiter, and Saturn — was already exceedingly well developed by the time of the astronomer Ptolemy around 150 a.d. By this we mean that the angular positions of these planets as a function of time had been catalogucd with remarkable accuracy and over a long enough span for their periodic returns to the same position in the sky to bc cxtremely well known. We have pointed out pre- viously, however, that the interpretation of such results depends on the model of the solar system that one uses. Let us now look more carefully at the original observational data and the con- clusions that can be drawn from them. The first thing to recognize is that, whether or not one accepts the earth as the real center of the universe, it is the center as far as all primary observations are concerned. From this vantage point, the motion of each planet can be described, to a first approximation, as a small circle (the epicycle) whose center moves around a larger circle (the deferent). Now there are some facts about the motions of two particular planets— Mercury and Venus— that point the way to some far-reaching conclusions. These are 1. That for these two planets, the time for the center C of the epicycle [Fig. 8-1 (a)] to travel once around the deferent is exactly 1 solar year — i.e., the same time that it takes the sun to complcte one circuit around the ecliptic. 2. The planets Mercury and Venus never get far from the sun. They are always found within a limited angular range from the line joining the earth to the sun (about ±22J° for Mercury, 246 Universal gravitation Fig. 8-1 (a) Motions of .'/.'(■ sun and Venus as seen from the earlh. Venus always lies within the angular range ±$ m of the sun' s direction. (6) Heliocentric picture of the same situation. ±46° for Venus). Both of these facts are beautifully accounted for if we go over to the heliocentric, Copernican system [Fig. 8— l(b)]. We see that the larger circle of Fig. 8-l(a) corresponds in this case to the earth's own orbit around the sun, of radius rs, and the smaller circle — the epicycle — represents the orbit of the other planet (Venus or Mercury, as the case may be). Given this interpretation, we can proceed to make quantitative in- ferences about the radii of the planetary orbits themselves. This is a crucial advance of the Copernican scheme over the Ptolemaic. Although Ptolemy had excellent data, they were for him just the source of purely geometric parameters, but with Copernicus we arrive at the basis of a truly physical model. Thus in Fig. 8-1 (b) the maximum angular deviation, d m , of the planet P from the earth-sun line ES defines the planet's orbit radius r by the equation — = sin i (r B > r) (8-la) The radius of the earth's orbit is clearly a natural unit for mea- suring other astronom ical distances, and has long been used for this purpose: 1 astronomical unit (AU) mean distance from earth to sun (1.496 X 10" m) In terms of this unit, we then have 247 The orbits of the planets For Mercury: r « sin 225° For Venus: r « sin 46° 0.38 AU 0.72 AU When it comes to the other planets (Mars, Jupiter, and Saturn) the tables are turned. These planets are not closely linked to the sun's position; they progress through the full 360° with respect to the earth-sun line. This can be readily explained if we interchange the roles of the two component circular motions, so that the large primary circle (the deferent) is taken to be the orbit of the planet, now larger than that of the earth, and the epicycle is seen as the expression of the earth 's orbit around the sun. In the case of Jupiter, for example, the Ptolemaic picture is represented by Fig. 8-2(a) and the Copernican picture by Fig. 8-2(b). Thus the periodic angular swing, ±0 m , of the epicycle is now related to the ratio of orbital radii through the equation ^=sin- r (r B < r) (8-1 b) in which the roles of r and r B are reversed with respect to Eq. (8-la). Ptolemy's recorded values of 6 m for Mars, Jupiter, and Saturn were about 41°, 1 1°, and 6°, respectively. These would then lead to the following results: For Mars: r « esc 41° « 1.5 AU For Jupiter: r « esc 11° « 5.2 AU For Saturn: r « esc 6° ~ 9.5 AU Thus with the Copernican seheme (and this was its great triumph) it became possible to use the long-established data to construct Fig. 8-2 (a) Motions of the sun and Jupiter as seen from the earth. The angle 8 m here charaeterizes the magnitude of the retrograde (epicyclic) motion. (6) Helio- centric picture of the same situation. 248 Universal gravitation Fig. 8-3 Universe according lo Copernicus. (Reproduced from his hisloric work, De Revolutionibus.) a picture of the planets in their orbits in order of their increasing distance from the sun. Figure 8-3 is a reproduction of the historic diagram by which Copernicus displayed the results in his book {De Revolutionibus) in 1543. The data with which Copernicus worked (and Ptolemy, too, 1400 years before him) were actually far too good to permit a simple picture of the planets describing circular paths at con- stant speed around a common center. Thus Copernicus carried out a detailed analysis to find out how far the center of the orbit of each planet was offset from the sun. But even with this adjustment, the detailed change with time of the angular positions of the planets could not be fitted unless the motion around the orbit was made nonuniform. Copernicus, like Ptolemy before him, introduced auxiliary circular motions to deal with the problem, but this, as we now know, was not the answer and we shall not discuss its compIexities. For the moment we shall use the basic idealization of uniform circular orbits and set aside until later the rcfincmcnts that were first mastered by Kepler when he recognized the planctary paths as being ellipses. PLANETARY PERIODS The problem of determining the periodicities of the planets, like that of finding the shapes of their orbits, must begin with what 249 Planctary pcriods Fig. &-4 (a) Relative posilions of the sun, the earth, and Jupiter at the beginning (SEiJi) and end (SE2J2) ofone synodic period. (b) Comparable dia- gram for the sun, the earth, and Venus, allowing for the fact that Venus must be offset from the line between sun and earth ifit is to be visible. E t fr y E, *£ 2 \ / \ / \ / 1 i v > / V/ 1 j 1 v, E x (a) (b) ^^HBH BI can be observed from the moving platform that is our earth. The recurring situation that can be most easily recognized is the one in which the sun, the earth, and another planet return, after some characteristic time, to the same positions relative to one another. The length of this recurrence time is known as the synodic period of the planet in question. In terms of a helio- centric model of the solar system, this is easily related to the true (sidereal) period of one complete orbit of the planet around the sun. Consider first the case of one of the outer planets, say Jupiter. Figure 8-4(a) shows a situation that can be observed from time to time. The positions of the sun, the earth, and Jupiter lie in a straight line. Observationally this could be established by finding the date on which Jupiter passes aeross the celestial meridian at midnight, thus placing it 180° away from the sun. ' Now if one such alignment is represented by the positions Ei and J t of the earth and Jupiter, the next one will oecur rather more than 1 ycar later, when the earth has gained one whole revolution on Jupiter. This is shown by the positions E 2 and J 2 . Jupiter has traveled through the angular distance 9 while the earth goes through 2k + 6. Both Ptolemy and Copernicus knew "The celestial meridian is the projection, on the celestial sphere, of a plane containing the earth's axis and the point on the earth's surface where the observer is located. It is thus a great cirele on the celestial sphere, running from north to south through the observer's zenith point, vertically above him. Noon is the instant at which the sun crosses this celestial meridian in its daily journey from cast to west. 250 Universal gravitation that the length of the synodic period separating these two con- figurations is close to 399 days. Let us denote the synodic period in general by the symbol r. Then if the earth makes ns complete revolutions per unit time and Jupiter makes nj revolutions per unit time, we have H£r = rtjr + 1 But n E and tij are the reciprocals of the periods of revolution T e and Tj of the two planets. Thus we have and solving this for Tj we have Tj = - T i. (8-2a) 1 - T e /t Putting Tb/t ~ 365/399 w 0.915, we thus find that The same type of observation and calculation can be applied to Mars and Saturn and the other outer planets that we now know. When we come to Venus and Mercury, however, the situation, as with the determination of orbital radii, is a little different. First is the practical difficulty that we cannot, at least with the naked eye, see these planets when they are in line with the sun, because it would rcquire looking directly toward the sun to do so. We can easily get around this by considering any other situation [see Fig. 8-4(b)] in which the angle between the direc- tions ES and EV is measured. This particular diagram shows Venus as a morning star, appearing above the horizon 1 hr or so before the sun as the earth rotates from west to east. The same value of the angle a will recur after one synodic period. This takes over l^yr — about 583 days, to be more precise. In this case, however, it is Venus that has gaincd one revolution on the earth. Thus instead of the form of the equation that applies to the outer planets, we now have tivr = ngt + 1 leading to the result 1 + Tb/t 251 Planetary periods Putting T E /r « 365/583 « 0.627, we find that 7V« ^»224 days It is a curious fact that Copernicus, in the introductory general account of his model of the solar system, quotes values of the planetary periods which are so rough that some of them could even be called wrong. These values are marked on his diagram (Fig. 8-3) and are repeated in his text: Saturn, 30 yr; Jupiter, 12yr; Mars, 2yr; Venus, 9 months; Mercury, 80 days. The worst cases are Mars (2 yr instead of about I i) and Venus (9 months instead of about 7$). This seems to havc led some peoplc to think that Copernicus had only a crude knowledge of the facts, which was certainly not the case. Perhaps he was careless about quoting the periods because his real interest was in the geometrical details of the planetary orbits and distances. The truth of the matter, in any event, is that his quantitative knowledge of both periods and radii, as spelled out in detail later in his book, was so good that the best modern values do not, for the most part, differ significantly from the ones he quoted. This is shown in Table 8-1, which lists both the Copern- ican and the modern data on the classical planets. (Incidentally, the values to be extracted from Ptolemy's data are almost iden- TABLE 8-1: DATA ON PLANETARY ORBITS Orbital radius, AU Synodic period, days Sidereal period Planet Copernicus Modern Copernicus Copernicus Modern Mercury 0.376 0.3871 115.88 87.97 days 87.97 days Venus 0.719 0.7233 538.92 224.70 days 224.70 days Earth 1.000 1.0000 — 365.26 days 365.26 days Mars 1.520 1.5237 779.04 1.882 yr 1.881 yr Jupiter 5.219 5.2028 398.96 11.87 yr 11.862 yr Saturn 9.174 9.5389 378.09 29.44 yr 29.457 yr tical with those of Copernicus, an astonishing tribute to those astronomers whose measurements, from about 750 b.c. up to the time of Ptolemy's own observations around 130 a.d., pro- vided the basis of his analysis.) KEPLER'S THIRD LAW The data of Table 8-1 point clearly to a systematic relationship between the planetary periods and distances. This is displayed 252 Universal gravitation Fig. 8-5 Smooth curue relating the periods and the orbital radii of the planets. graphically in Fig. 8-5. The precise form of the relationship was first discovered by Johann Kepler in 1618 and published by him the following year in his book The Harmonies of the World. In it he triumphantly wrote: "I first believed I was dreaming . . . But it is absolutely certain and exact that the ratio which exists between the periodic times of any two planets is precisely the ratio of the ^th powers of the mean distances . . . ." Table 8-2 TABLE 8-2: KEPLER'S THIRD LAW Planet Radius r of orbit of planet, AU Period T, r s /T 2 , days (AU)' i /(dayy 2 X 10" Mercury 0.389 87.77 7.64 Venus 0.724 224.70 7.52 Earth 1.000 365.25 7.50 Mars 1.524 686.98 7.50 Jupiter 5.200 4,332.62 7.49 Saturn 9.510 10,759.20 7.43 shows the data used by Kepler and a test of the near constaney of the ratio r 3 /T 2 . Figure 8-6 is a different presentation of the planetary data (actually in this case the data of Copernicus from Table 8-1) plotted in modern fashion on log-log graph paper so as to show this relationship: T~ r 3 ' 2 (8-3) This is known as Kepler's third law, having been preceded, 10 years earlier, by the statement of his two great discoveries (quoted 253 Kepler's third law !|HBHnaHMMngHBmHHn|HmaHgnBMjfl Fig. 8-6 Log-log plot of planetary period T versus orbit radius r, using data guoted by Copernicus. The graph shows that T is proporlional to r 3 '" (Kepler's third law). in the Prologue) concerning the elliptical paths of the individual planets. The dynamical expIanation of Kepler's third law had to await Newton's discussion of such problems in the Principia. A very simple analysis of it is possible if we again use the sim- plified picture of the planetary orbits as circles with the sun at the center. It then becomes apparent that Eq. (8-3) implies that an inverse-square law of force is at work. For in a circular orbit of radius r we have Expressing v in terms of the known quantities, r and T, we have v = 2itr 254 Universal gravitation 4 Z a r = ==- (toward the center) (8-4) From Newton's law, then, we infer that the force on a mass in a circular orbit must be given by F r =ma r =- *&£ (8-5) From Kepler's third law, however, we have the relation Y 2 = K (8-6) where K might be called Kepler's constant — the same value of it applies to all the planets traveling around the sun. From Eq. (8-6) we thus have l/T 2 = K/r z , and substituting this in Eq. (8-5) gives us Fr = - ^ (8-7) The implication of Kepler's third law, therefore, when analyzed in terms of Newton's dynamics, is that the force on a planet is proportional to its inertial mass m and inversely pro- portional to the square of its distance from the sun. Newton's contemporaries Halley, Hooke, and Huygens all appear to have arrived at some kind of formulation of an inverse-square law in the planetary problem, although Newton's, in terms of his deflnite concept of forces acting on masses, seems to have been the most clear-cut. The general idea of an influence falling off as l/r 2 was probably not a great novelty, for it is the most natural-seeming of all conceivable effects — something spreading out and having to cover spheres of larger and larger area, in proportion to r 2 , so that the intensity (as with light from a source) gets weaker according to an inverse-square relationship. The proportionality of the force to the attracted mass, as required by Eq. (8-7), was a feature of which only Newton appreciated the full significance. With his grasp of the concept of intcractions exerted mutually between pairs of objects, Newton saw that the reciprocity in the gravitational interaction must mean that the force is proportional to the mass of the attracting object just as it is to the mass of the attracted. Each object is the attracting agent as far as the other one is concerned. Hence the magnitudo of the force exerted on either one of a mutually 255 Kepler's third law gravitating pair of particles must be expressed in the famous mathematical statement of universal gravitation : F _ _ Gmxm2 (8 _ g) where G is a constant to be found by experiment, and m, and m 2 are the inertial masses of the particles. We shall return to the matter of determining G in practice, but first we shall discuss the famous problem that led Newton toward some of his greatest discoveries concerning gravitation. THE MOON AND THE APPLE It is an old story, but still an enthralling one, of how Newton, as a young man of 23, came to think about the motion of the moon in a way that nobody had ever done before. The path of the moon through space, as referred to the "fixed" stars, is a line of varying curvature (always, however, bcnding toward the sun), which crosses and recrosses the earth's orbit. But of course there is a much more striking way of looking at it — the familiar earth-centered view, which shows the moon describing an ap- proximatcly circular orbit around the earth. To this extent it is quitc like the planetary-orbit problem that we have just been discussing. But Newton, with his extraordinary insight, con- structed an intellectual bridge between this motion and the behavior of falling objects— the latter being such a commonplace phenomenon that it needed a genius to rccognize its relevance. He saw the moon as being just an object falling toward the earth like any other— as, for example, an apple dropping off a tree in his garden. A very special case, to be sure, because the moon was so much farther away than any other falling object in our ex- perience. But perhaps i t was all part of the same pattern. As Newton himself described it, ' he began in 1665 or 1666 to think of the earth's gravity as extending out to the moon's orbit, with an inverse-square relationship already suggested by Kepler's third law. We could of course just restate the cen- tripetal acceleration formula and apply it to the moon, but it is illuminating to trace the course of Newton's own way of dis- cussing the problem. In effect hc said this: Imagine the moon at any point A in its orbit (Fig. 8-7). If freed of all forces, it 'See the Prologue of this book. 256 Universal gravitation Fig. 8-7 Geometry ofa small portion ofa circular orbit, show- ing the deviation y from the tangential straiglu-line displace- ment AB ( = x) lltat would be followed in the absence of gracity. would travel along a straight line AB, tangent to the orbit at A. Instead, it follows the arc AP. If O is the center of the earth, the moon has in effect "fallen" the distance BP toward O, even though its radial distance r is unehanged. Let us caleulate how far the moon falls, in this sense, in 1 sec, and compare it with the distance of about 16 ft that an object projected horizontally near the earth's surface would fail in that same time. First, a bit of analytic geometry. If we denote the distance AB as x, and the distance BP as y, it will be an exceedingly good approximation to put y ~2r (8-9) One way of obtaining this result is to consider the right triangle ONP, in which we have ON = r - y NP = x OP = r Hence, by Pythagoras' theorem, (r - y) 2 + x 2 = r 2 x 2 = 2ry - y 2 Since y « r for any small value of the angle d, Eq. (8-9) follows as a good approximation. Furthermore, since (again for small d) the arc length AP (= s) is almost equal to the distance A B, we can equally well put 257 The moon and the apple y « *- (8-10) In order to put numbers into this formula we need to know both the radius and the period of the moon's orbital motion. The distance to the moon, as known to Newton, depended on the two-step process devised by the ancient astronomers— finding the earth's radius and finding the moon's distance as a multiple of the earth's radius. A rcminder of these classic measurements is given in Figs. 8-8 and 8-9 and the accompanying discussion (pp. 259-261). The final result, familiar to everyone, is that the moon's orbit radius r is about 240,000 miles ~ 3.8 X 10 8 m. Its period T is 27.3 days « 2.4 X 10 fi sec. Therefore, in 1 sec it travels a distance along its orbit given by r , s 2ttX 3.8 X 1 0* _ |WM (m 1 sec) s = 2A x 1Q| - - « 1000 m During this same time it falls a vertical distance, which we will denote y>i to identify it, given [via Eq. (8-10)] by (ini sec) >-i«7: 6 -^lo« =1 - 3X,0 ~ 3m In other words, in 1 sec, while traveling "horizontally" through a distance of 1 km, the moon falls vertically through just over 1 mm, or about 5 V in.; its deviation from a straight-line path is indeed slight. On the other hand, for an object near the earth's surface, projected horizontally, the vertical displacement in 1 sec is given by y2 = yr- = 4.9 m Thus 'i - « x 'O"'' - m Newton knew that the radius of the moon's orbit was about 60 times the radius of the earth itself, as the ancient Greeks had first shown. And with an inverse-square law, if it applied equally well at all radial distances from the earth's center, we would expect yjy* to be about 1/3600. It must bc right! And yet, what an astounding result. Even granted an inverse- square law of attraction between objccts separated by many times their diamctcrs, one still has the task of proving that an 258 Universal gravitation object a few feet above the earth's surface is attracted as though the whole mass of the earth were concentrated at a point 4000 miles below the ground. Newton did not prove this result until 1685, nearly 20 years after his first great insight into the problem. He published nothing, either, until it all came out, perfect and complete, in the Principia in 1687. One way of solving the problem follows on p. 262 (after the special section below). FIND1NG THE D1STANCE TO THE MOON The earth's radius About 225 B. c. Eratosthenes, who lived and worked at Alexandria near the mouth of the Nile, reported on measurements made on the shadows cast by the sun at noon on midsummer day. At Alexandria (marked A in Fig. 8-8) the sun's rays made an angle of 7.2° to the local vertical, whereas corresponding measurements made 500 miles farther south at Syene (now the site of the Aswan Dam) showed the sun to be exactly overhead at noon. (In other words, Syene lay almost exactly on the Tropic of Cancer.) It follows at once from these figures that the arc AS, of length 500 miles, subtends an angle of 7.2° or g rad at the center of the earth. Hcnce 500 _ 1 Re ~ 8 or Re ~ 4000 miles Fig. 8-8 Basis of the method useci by Eratosthenes to find the earth's radius. When the midday sun was exactly overhead at Syene (S) its rays fell at 7.2° to the verti- cal at Alexandria (A). 259 Finding the distance lo the moon The moon's distance measured in earth radii Hipparchus, a Grcek astronomer who lived mostly on the island of Rhodes, made observations in about 130 B.c. from which he obtained a remarkably accurate estimate of the moon's distance. His method was one suggested by another great astronomer, Aristarchus, about 150 years earlier. The method involves a clear understanding of the positional relationships of sun, earth, and moon. We know that sun and moon subtend almost exactly the same angle a. at the earth. Hipparchus measured this angle to be 0.553° (« 1/103.5 rad); he also knew what Aristarchus before him had found — that the sun is far more distant than the moon. Hipparchus used this knowledge in an analysis of an eclipse of the moon by the earth (Fig. 8-9). The shaded region indicates the area that is in com- plete shadow; its boundary lines PA and QB make an angle a with one another, because this is the angle between rays coming from the extreme edges of the sun. The moon passes through the shadowed region, and from the measured time that this passage took, Hipparchus deduced that the angle subtended at the earth by the arc BA was 2.5 times that subtended by the moon itself. Thus Z AOB « 2.5a. Let us now do some geometry. If the distance from the earth's center to the moon is D, the length of the arc BA is very nearly equal to the earth's diameter PQ diminished by the amount aD: AB « 2R B - aD Fig. 8-9 Basis of the method used by Hipparchus tofind the moon's distance. The method depended on obseroing the duration {and hence the augular widtli) of the moon's total eclipse in the shadow of the earth, as represented by the arc AB. 260 Universal gravitation But we also have AB _ D ~ 2.5a or AB « 2.5aZ> Substituting this in the first equation WC have 3.5aD i» 2R F , or D _ 2 3.5o Since a «s 1/103.5 rad, this gives Z) _ 5- Combining this with the value of .Re itself, we have D » 236,000 miles Modern methods Refined triangulation techniques give a mean value of 3,422.6", or 0.951°, for the angle subtended at the moon by the earth's radius. Using the modern value of the earth's radius (Re = 6378 km = 3986 miles) one obtains almost exactly 240,000 miles for the moon's mean distance. Such traditional methods, however, are far surpassed by the technique of making a precision measurement of the time for a radar echo or laser reflection to return to earth. The fiight time of such signals (only about 2.5 sec for the roundtrip) can be measured to a fraction of a microsecond, giving range deter- minations that are not only of unprecedcntcd accuracy but are also effectively instantaneous. THE GRAVITATIONAL ATTRACTION OF A LARGE SPHERE It has long been suggested that Newton's failure to publicize his discovery about an invcrse-square law of the earth's gravity 261 The gravitational attraction of a large sphere extcnding to the moon was due primarily tO an actual numerical discrepancy, resulting from his use of an erroneous value for the earth's radius. This would, via Eq. (8-10), falsify the value of the moon's distance of fail, since /• (the radius of the moon's orbit) was calculated, according to the method discovered by Hipparchus, in tcrms of the earth's radius. When Newton first did the calculation he was home in the countryside, out of reach of reference books, and it is reliably recorded that he calculated the earth's radius by assuming that 1° of latitude is 60 miles, instead of the correct figurc of nearly 70 miles. Be this as it may, it remains almost certain that Newton, with his outstandingly thorough and critical approach to problems, would never have regarded the theory as complete until he had solved the problem of gravitation by large objects. Let us now consider a way of analyzing this problem. (In Chapter 1 1 we shall tackle it again in a more sophisticated way.) Suppose we have a large solid sphere, of radius R , as shown in Fig. 8-10(a), and wish to calculate the force with which it at- tracts a small object of mass m at an arbitrary point P. We shall assume that the density of the material of the sphere may vary with distance from the ccnter (as is the case for the earth, to a very marked degree) but that the density is the same at all points equidistant from the center. We can then consider the solid sphere to be built up of a very large number of thin uniform Fig. 8-10 (a) A solid sphere can be regarded as built up ofa set ofthin concentric splierieal shells. (b) The gravitational effect of an individual shell can befound by treating it as an assembtage ofeireular zones. 262 Universal gravitation spherical shells, like thc successive layers of an onion. The total gravitational effect of the sphere can be calculated as the super- position of the effects of all these individual shells. Thus the basic problem becomes that of calculating the force exerted by a thin spherical shell of arbitrary radius, assuming that the funda- mental law of force is that of the inverse square between point masses. In Fig. 8-10(b) we show a shell of mass M, radius R, of negligible thickness, with a particle of mass m at a distance r from the center of the shell. If we consider a small piece of the shell, near point A, the force that it exerts on m is along the line AP. It is clear from the symmetry of the system, howcver, that the resultant force due to the whole shell must be along the line OP; any component of force transverse to OP due to material near A will be canceled by an equal and opposite contribution from material near A'. Thus if we have an element of mass dM near A, we need only consider its contribution to the force along OP, i. e., the radial direction from the center of the shell to m. Hence we have GmdM s 2 Let us now consider a complete belt or zone of the shell, shown shaded in the diagram. It represents the portion of the shell that is contained between the directions and 6 + dd to the axis OP, and the same mean values of s and <p apply to every part of it. Thus, if we calculate its mass, we can substitute this value as dM in Eq. (8-11) to obtain the contribution of the whole belt to the resultant gravitational force along OP. Now the width of the belt is R dd and its circumference is 2irR sin 9; thus its area is 2irR 2 sin 6 dd. The area of the whole shell is 4irR 2 ; hence the mass of the belt is given by _. 2TrR 2 sm6d6 .. M . Q „ dM = — — M = — sin d6 Thus Eq. (8-11) gives us GMm co s f sin 9 dd .. ,. OFr = j Ti (o-lZ) Our task now is to sum the contributions such as dF r over the 263 The gravitational attraction of a largc sphere whole of the shell, i.e., over thc whole range of values of .?, <p, and e. This looks like a formidable task, but with the help of a little calculus (another of Newton's inventions!) the solution turns out to be surprisingly straightforward. From the geometry of the situation [Fig. 8-10(b)], it is possible to exprcss both of the angles and <p in terms of two fixed distances, r and R, and the variable distance s. By two separate applications of the cosine rule we have ,»+ *»- l r 2 + s 2 - R 2 cose = - -uar "* v =~ ~^T From the first of these, by differentiation, we have • „ , n sds sin 6 dd = — -- rR Hence, substituting the values of cos <p and sin 6 dd in Eq. (8-12), we obtain GMm (r 2 + s 2 - R 2 )ds dFr = " ~4^R ^ The total force is found by integrating this expression from the minimum value of s (= r — R) to its maximum value (r + R). Thus we have "--^ir d± 4^- 'f - The integral is just the sum of two elementary forms; we have r 2 -R 2 Inserting the limits, we then find that / r+B 2 J. 2 _ » 2 f "*"% — ds m [(r + R)- (r - R)] T-R S* _ ( r 2 - R 2 _ r 2 - R 2 \ \r+ R " r- R J = 2R - (r- R) + (r + R) = 4R 264 Universal gravitation Substituting this value of the definite integral in Eq. (8-13) we have F r = - ^ (8-14) What a wonderful result! It is of extraordinary simplicity, and the radius R of the shell does not appear at all. It is uniquely a consequence of an inverse-square law of force between particles; no other force law would yield such a simple result for the net effect of an extended spherical object. Once we have Eq. (8-14), the total effect of a solid sphere follows at once. Regardless of the particular way in which the density varies between the center and the surface (provided that it depends only on R) the complete sphere does indeed act as though its total mass were concentrated at its center. It does not matter how close the attracted particle P is to the surface of the sphere, as long as it is in fact outside. Take a moment to consider what a truly remarkable result this is. Ask yourself: Is it obvious that an object a few feet above the apparently fiat ground should be attracted as though the whole mass of the earth (all 6,000,000,000,000,000,000,000 tons of it!) were con- centrated at a point (the earth's center) 4000 miles down? It is about as far from obvious as could be, and there can be little doubt that Newton had to convince himself of this result before hc could establish, to his own satisfaction, the grand connection between terrestrial gravity and the motion of the moon and other celestial objects. OTHER SATELLITES OF THE EARTH Newton's thinking quite explicitly embraced the possibility — at least theoretically — of having other satellites of the earth. Figure 8—1 1 is an illustration from Newton's book, The System of the World (which is incorporated in the Principia); it shows the transition from the effectively parabolic trajectories of short- range projectiles (although the apparent parabolas are really small parts of ellipses) to a perfectly cireular orbit and then to other elliptic orbits of arbitrary dimensions. Let us derive the formulas for the required velocity v and the period T of a satellite launehed horizontally in a cireular orbit at a distance r from the center of the earth. The necessary 265 Other satellites of the earth Fig. 8-11 Newlon's diagram showing the transit ion front normal parabolic trajectories to complete orbits encircling the earth. (From The System of the World.) force to maintain circular motion is provided by gravitational attraction: mv r = G Mnm r* where Mg is the mass of the earth, m the mass of the satellite, and G the universal gravitational constant. Solving for v, .-(«T (8-15) It is often convenient to express this result in terms of more familiar quantities. We can do this by noticing that, for an object of mass m at the earth 's surface, the gravitational force on it, by Eq. (8-8), is F. - GMatn Re 2 But this is the force that can be set equal to mg for the mass in question. Hence we have mg = GMgm Rs 2 or GM E = gR E Substituting this in Eq. (8-15), we get 266 Universal gravitation v 1/2 -W The period, T, of the satellite is then given by 2xr _ 2^ /2 v g ll2 R E (8-16) Putting g = 9.8 m/sec 2 , /?g = 6.4 X 10° m, we have a nu- merical formula for the period of any satellite in a circular orbit of radius r around the earth: (Earth satellites) T w 3.14 X 10~ 7 r 3/2 (8-17) where T is in seconds and r in meters. For example, a satellite at minimum practicable altitude (about 200 km, say), has r « 6.6 X 10° m, and hence T « 5.3 X 10 ;i sec w 90 min The first man-made satellite, Sputnik I (October 1957) had an orbit as shown in Fig. 8-1 2(a). Its maximum and minimum distances from the earth's surface were initially 228 and 947 km, respectively, giving a mean value of r equal to about 6950 km. Fig. 8-12 (a) Orbit of Sputnik I, the first man-made satellite (October 1957). (A) Synchronous satellite com- munication System. Orbital diameter in relation to earth's diameter is approximately to scale. 267 Othcr satellites of the earth With this value of r, Eq. (8-17) gives an orbital period of about 96 min, which agrees closely with the observed figure. Particular interest attaches to synchronous satellites that have an orbital period equal to the period of the earth's rotation on its axis. If placed in orbit in the earth's equatorial plane, such satellites will remain above the same spot on the earth's surface, and a set of three of them, ideally in a regular triangular array as shown in Fig. 8-12(b), can provide the basis of a worldwide Communications system with no blind spots. Putting T = 1 day in Eq. (8-17), one finds r « 42,000 km or 26,000 miles. Thus such satellites must be about 22,000 miles above the earth's sur- face, i. e., about 5£ earth radii overhead. The first such satellite to be successfully launched was Syncom II in July 1963. Equation (8-16), on which the above calculations are based, has a very noteworthy feature. A satellite traveling in a circular orbit of a given radius has a period independent of the mass of the satellite. Thus a massive spaceship of many tons will, for the same value of r, have precisely the same orbital period as a flimsy object such as one of the Echo balloons, with a mass of only about 100 kg — or, for that matter, a small piece of interplanetary debris with a mass of only a few kilograms. This result is a direct consequence of the fact that the gravitational force on any object is strictly proportional to its own mass. THE VALUE OF G, AND THE MASS OF THE EARTH Although the result expressed by Eq. (8-14) was obtained by considering a large sphere and a small particle, one can quickly convince oneself that it is also the correct statement of the force between any two spherical objects whose centers are a distance r apart. For suppose that we have two such spheres, as shown in Fig. 8-1 3(a). The calculation that we have carried out shows that one sphere (say the one on the left) attracts every particle of the other as if the left-hand mass were a point [Fig. 8-1 3(b)]. This therefore reduces the problem to the mutual gravitational attraction between a sphere (the right-hand sphere, of mass m) and a point particle of mass M. But now we can apply the result of the Iast section a second time. Thus we arrive at Fig. 8-1 3(c), with two point masses separated by a distance r, as a rigorously correct basis for calculating the force of attraction between the two extended masses shown in Fig. 8-1 3(a). The above result is important in the analysis of the experi- ment, already described in Chapter 5, for finding the universal 268 Universal gravitation Fig. 8-13 (a) Two gracilaiing spheres at small separa- lion. (b) Effect ofone spliere (M) can be calculated by treating it as a poinl mass. (c) The argumen! can be repeated, so thal the attraction between the spheres can be calculated as lhougli bolh were poinl masses. gravitation constant. G, from the measured force between two spheres of known masses. In order to get the biggest possible effect with an interaction that is so extremely weak, it is usual to arrange things so that the centers of the spheres are separated by only a little more than the sum of the radii. It is then a great convenience to be able to calculate the force, even under these conditions, on the basis of Eq. (8-14). Notice, however, that the result holds only for spheres. Some of the measurements to determine G have made use of cylindrical masses, because of the greater ease of machining them to high precision. In such cases it becomes necessary to calculate the net force by an explicit integration over the spatial distribution of material. The prescntly accepted value of G, as obtained from labora- tory measurements of the force exerted between two known masses, is (as already quoted in Chapter 5): G = 6.670 X 10-" m : 7kg-sec 2 (8-18) Newton himself did not know the value of G, although he made a celebrated guess a t the mean density of the earth, from which he could have obtained a conjectural figure. In Book III of the Principia, he remarks at one point as follows : "Since ... the common matter of our earth on the surface thereof is about twice as heavy as water, and a little lower, in mines, is found about three, or four, or even five times heavier, it is probable that the quantity of the whole matter of the earth may be five or six times greater than if it consisted all of water . . . ." If we denote the mean density of the earth as p and its radius as R, the gravitational force exerted on a partiele of mass m just at the earth's surface is given by 269 The value of G', and ihc mass of ihc earth F = 9Mn (8-19) R 2 where Hence 4* F = -- (GpR)m Since, however, this is just the force that gives the particle an acceleration g in free fail, we also have F = mg It follows, then, that g = ^GpR (8-20) If in this cquation we put g « 9.8 m/sec 2 , R ~ 6.37 X 10° m, and (using Newton's estimate) p ~ 5000 to 6000 kg/m 3 , we find that G « (6.7 ± 0.6) X 10-" m :, /kg-sec 2 Thus Newton's estimate was almost exactly on target. In prac- tice, of course, the calculation is done the other way around. Given the directly determined value of G [Eq. (8-18)] we sub- stitute in Eqs. (8-19) and (8-20) to find the mass and the mean density of the earth. The result of these substitutions (with R = 6.37 X 10° m) is M = 5.97 X 10 24 kg p = 5.52 X 10 3 kg/m 3 LOCAL VARIATIONS OF g If we take the idealization of a perfectly spherical, symmetrical earth, then the gravitational force on an object of mass m at a distance h above the surface is given by GMm F = (R + A)2 270 Universal gravitaiion If we identify F with m times the value of g at the point in ques- tion, we have ■» - m (8 - 2,) For /i « R, this would imply an almost exactly linear decrease of g with height. Using the binomial theorem, we can rewrite Eq. (8-21) as follows: ,,, GM(.,h\— m \ * * 2 ~ 7 Hence, for small h, we have gCA)-w(l-f) (8-22) where g = GM/R 2 , the value of g at points extremely close to the earth's surface. [Alternatively, we can use a calculus method that can be extremely useful whenever we want to consider the fractional variation of a quantity. It is based on the fact that the differentiation of the natural logarithm of a quantity leads at once to the fractional variation. In the present case we have .. GM «W = -^T Therefore, ln g = const. — 2 ln r Differentiating, As k _ 2 Ar g r Hence, putting r = R, g = go, and Ar = h, we have Ag ^ - 2go - which leads us back to Eq. (8-22). Notice how this method frees us of the need to concern ourselves with the values of any multiplicative constants that appear in the original equation — e.g., the value of GM in the equation for g. A recognition of this fact can enable one to avoid a lot of unnecessary arithmetic in 271 Local variations ofg the computation of small changes of one quantity that depends upon another according to some well-defined functional re- lationship.] Newton's contcmporary, Robert Hooke, made several efforts to detect a variation of the gravitational attraction with height. He did this by looking for any changes in the measured weights of objects at the tops of church towers and the bottoms of deep wells. Not surprisingly, he was unable to find any dif- ference. By Eq. (8-22) one would have to ascend to a point about 1000 ft above ground (e. g., the top of the Empire State Building) before the decrease of g was even as great as I part in 10,000. As we shall see in a moment, however, such variations are detected with the greatest of ease by modern techniques. Superimposed on the systematic variations of the gravita- tional force with height are the variations produced by in- homogeneities in the material of the earth's crust. For example, if one is standing above a subterranean deposit of salt or sand, much lower in density than ordinary rocks, one would expect the value of g to be reduced. Such changes, although extremely small, can be measured with remarkable accuracy by modern instruments and have become a very valuable tool in geophysical prospccting. Almost all modern gravity meters are static instruments, in which a mass is in equilibrium under the combined action of gravity and an elastic restoring force supplied by a spring. In other words, it is just a very sensitive spring balance. A change in g as the instrument is moved from one point to another leads to a minute change in the equilibrium position, and this is de- tected by sensitive optical methods or electrically by, for ex- ample, making the suspended mass part of a tuned circuit whose capacitance, and hencc frequency, is changed by the slight displaccment. To be useful, such instruments must be capable of detecting fractional changes of g of 10 -7 or less. Figure 8-14(a) shows the basic construetion of one such device. With it one can trace out contours of constant g over a region of interest. Figure 8-14(b) shows the results of such a survey, after allowance has been made for effects due to varying altitude, surface features, and so on. Such contours can give good in- dications of ore concentrations. The primary unit of mcasure- ment in these gravity surveys is known as the gal (after Galileo): 1 gal = 1 cm/sec 2 « \Q~* g 272 Universal gra vital ion (a) Fig. 8-14 ia) Sketch of basic features ofa sensitiue grammeter, made offused quarlz. The arm marked W acts as the main weight. U is picoied al A and B and carries a pointer P. The restoring force is provided by a control spring S\ and a null reading can be oblained wilh the help ofthe ca/ibrajed spring 5 2 . (.b) Example ofa gravity surcey ouer an area about 400 by 500 m, with contours of conslant g indicating an ore deposit. (After a survey made by the Boliden Mining Co., Sweden, and reproduced in D. S. Parasnis, Mining Geophysics, Elsevier, Amsterdam, 1960.) 273 Local varialions of g This is far too large for convenience, so most surveys, like that of Fig. 8-14(b), show contours labeled in terms of milligals (1 mgal = 10 -3 cmsec 2 « 10~"g). Under the best conditions, relativc mcasurements accurate to 0.01 mgal may be achievable. One can appreciate how impressive this is by noting that a change of g by 0.01 mgal (1 part in 10 8 ) corresponds to a change in elevation of only about 3 cm at the earth's surface! THE MASS OF THE SUN Let us return to the simple picture of the solar system in which each planet deseribes a cireular orbit about a fixed central sun (Fig. 8-15). We have seen, in the diseussion of Kepler's third law, how the use of Newton's Iaw of motion implies that the foree on the planet is given, in terms of its mass, orbital radius, and period, by the following equation [Eq. (8-5)]: F r = - . 2 4tt mr /-■ According to the basic law of the foree, however, as exprcssed by Newton's law of universal gravitation [Eq. (8-8)], the value of F r is given by the equation F r = - GMm r* where M is the mass of the sun. From the equality of these two expressions, we obtain the following result: 4*y GM (8-23) We may again note the feature, already commented on in con- Fig. 8-15 Planelary orbit approxi- maled by a cirele wiih the sun at the cemer. 274 l Iniversal gravitation nection with earth satellites, that the period is independent of the mass of the orbiting object, in this case the earth itself or some other planet. What does matter is the mass of the sun, and if we turn Eq. (8-23) around, we have an equation that tells us the value of this mass, M, in terms of observable quantities: *-££ <H4> Kepler's third law expresses the fact that the value of r 3 /T 2 is indeed the same for a U the planets. The statement of this result does not, however, rcquire the use of absolute values of r — or, for that matter, of T either. It is sufficient to know the values of r and T for the various planets as multiples or fractions of the earth's orbital radius and period. In order to deduce the mass of the sun from Eq. (8-24), however, the use of absolute values is essential. We have seen, earlier in this chapter, how the length of the earth's year has been known with great accuracy since the days of antiquity. A knowledge of the distance from the earth to the sun is, however, rather recent. The development of this knowledge makes an interesting story, which is summarized in the special seetion following. The final result, expressed as a mean distance in meters, can be substituted as the value of r in Eq. (8-24), along with the other necessary quantities as folio ws: r*** 1.50 XlO H m T K ~ 3.17 X 10 7 sec G = 6.67 X 10-" m 3 /kg-sec 2 We then find that M„ m «■ 2.0 X 10 30 kg FINDING THE DISTANCE TO THE SUN The first attempt to estimate the distance of the sun was made by the great Greek astronomer, Aristarchus, in the third century B.c, and he arrived at a result which, although quite erroneous, held the field for many centuries. His method, sound in principle but made ineffectual by the great remoteness of the sun, is in- dicated in Fig. 8-16(a). He knew that one half of the moon was illuminated by the sun and that the phases of the moon were the result of viewing this illuminated hemisphere from the earth. 275 Findina the distance to ihc sun Sun Earth Fig. 8-16 (a) Melhod allempied by Arislarchus to find Ihe sun's dislance by measuring ihe angle SEM at half- moon. (b) Triangutation melhod of cslablishing the scale of ihe solar sysiem by finding ihe dislance o/ Mars, iising Ihe earth' s radius as a base line. (c) Direel delerminalion of ihe sun's dislance by obserc'mg Ihe transit of Venus front differenl poinls on Ihe earth. When the moon is exactly half full, the angle SME is 90°. If, in this situation, an exact measurement can bc made of the angle d, the difference in directions of the sun and moon as seen from a point on the earth, we can deduce the angle a (= 90° - 6) subtcnded at the sun by the earth-moon distance r m- Aristarchus judged to be about 87°, which gives a ~ 3° ~ 55 rad and hcnce /\s « 20/\u. Since, however, the measured angle is 9 and not a, the error in the final result may be (and is) very great. Our present knowledge tells us that the value of in the situation represented by Fig. 8-16(a) is actually about 89.8° instead of 87°; this relatively small change in 6 raises the ratio r s /r M to several hundred instead of 20. A complctcly different attack on the problem was initiated by Kcpler, although its full exploitation was not possible until much later. Even so it at once becamc clear that the sun is more distant than Aristachus had concluded. The basis of the method is indicated in Fig. 8-16(b). It involves observations on 276 Universal gravitation the planet Mars. When Mars is closest to the earth, it lies on a line joining both planets to the sun. Under these conditions the distance between them is the difference between their orbital radii. Now if Mars is viewed from two different points on the earth, it should appear in slightly different directions with respect to the vastly more distant background of "fixed" stars. The particular angular difference, S, for observers placed at A and B is called the parallax; it is the angle subtended by the earth's radius at the position of Mars. To measure this angle one does not need to have observers at two different points on the earth; the rotation of the earth itself would carry an observer from A. to B in about 6 hr during a given night. Now Kepler was able to deduce from the very careful observations of his master, Tycho Brahe, that the value of S must be less than 3 minutes of arc, which is about 1/1200 rad; he could conclude that the distance to Mars in this configuration must be greater than 1200 earth radii or about 5 million miles. Then, using the known relatioe values of orbital radii from the Copernican seheme (Table 8-1), it follows that the distance of the sun from the earth is more than 2400 earth radii, i.e., more than 10 million miles. John Flamsteed, a contemporary of Newton to whose observations Newton owed a great deal (he was Astronomer Royal from 1675 to 1720), reduced the upper limit on the parallax of Mars to about 25 seconds of arc, and concluded that the sun's distance was at least 80 million miles. An Italian astronomer, Cassini, arrived at a specific value of about 87 million miles at about the same time, using observations made by himself in Europe and by a French astronomer, Richer, at Cayenne in South America. Another contemporary of Newton's — Edmund Halley 1 — proposed a method that finally led, 100 years later, to the first precision measurements of the sun's distance. The method involved what is known as a transit of Venus, i.e., a passage of Venus aeross the sun's disk, as seen from the earth. Figure 8-16(c) illustrates the basis of the method. As it passes aeross the sun, Venus looks like a small black dot. Its apparent path, and also the times at which the transit begins or ends, depend on the position of the observer on earth. Since the motion of Venus ■Edmund Halley, best known for the comet named after him, succeeded Flamsteed in 1720 as Astronomer Royal. Long before this, however, he had been very active in physics and astronomy. He became a devoted friend and admirer of Newton, and it was largely through his help and persuasion that the Principia was published, 277 Finding the distance to the sun is accurately known, the timing of the transit can be used to yield accurate measures of the differences in angular positions of Venus as seen by observers at different positions. From such observations the parallax of Venus can be inferred, after which one can use an analysis just like that for Mars. These transits arc fairly rare, because the orbits of the earth and Venus are not in the same plane, but Halley pointed out that a pair of them would occur in 1761 and 1769, and again in 1874 and 1882, and then in 2004 and 2012. From the first two of these (both occurring long after Halley 's own death) the solar parallax was found to be dcfinitely between about 8.5 and 9.0 seconds of arc, correspond- ing to a distance of between about 92 and 97 million miles. Thus the currently accepted result was approached. (The best mea- surements of this type have been made on the asteroid Eros at its elosest approach to the earth.) Further refinements came with the observations made in the late nineteenth century. One of the most notable of these was the use of an accurately known value of the speed of light to deduce the diameter of the earth's orbit from the accumulated time lag, over a period of 6 months, in the observed eclipses of the moons of Jupiter. The situation is indicated in Fig. 8-17. While the earth moves from E x to E 2 , Jupiter moves only from J, to J \. This introduces an extra transit time of about 16 min for the light that tells us that one of Jupiter's moons has, for example, just appeared from behind the planet. Knowing that the speed of light is 186,000 miles/sec in empty space, we can deduce that the earth's orbital radius is equal to this speed times about 480 sec, or about 90 million miles. (The caleulation was originally donc just the other way around, by the Danish astron- omer Roemer in 1675. Using an approximate value of the distance from earth to sun, he made the first quantitative estimate of the speed of light.) Fig. 8-17 Measure- ment of the diameter of the earth's orbit by observing the eclipses of Jupiter's moons and the apparent delays due to the travel time of light through space. 278 Universal gravitation Although the modern measurements of the sun's distance are of great accuraey, we must still reckon with the fact that this distance varies diiring the course of a year. If we ignore this relatively small variation, however, we can make use of the average value, already quoted near the beginning of this chapter: r E = 1 AU = 1.496 X 10 n m MASS AND WEIGHT Perhaps the most profound contribution that Newton made to science was the fundamental connection that he recognized be- tween the inertial mass of an object and the earth's gravitational force on it — a force roughly equal to the measured weight of the object. (Remember, we have defined weight as the magnitude of the force, as measured for example on a spring balance, that holds the object at rest relative to the earth's surface.) It had been known since Galileo's time that all objects near the earth fail with about the same acceleration, g. Until Newton it was just a kinematic fact. But in terms of Newton's law it took on a much deeper significance. If an object is observed to have this acceleration, there must be a force F„ acting on it given by F = ma, i.e., F, = mg (8-25) It then becomes a vitally significant dynamical fact that, since the acceleration g is the same for all objects, the force F, causing it is strictly proportional to the inertial mass. To appreciate how remarkable this result is, imagine starting from serateh to in- vestigate the force of attraction between two objects in a purely static experiment. One measures the force by balancing it with a springlike device — a torsion fiber. One finds a quantity, which might be called (by analogy with electrical interaetions) a gravita- tional charge, q g . This "charge" is charaeteristie of any object and has, as far as these experiments are concerned, nothing at all to do with the inertial mass, which is defined solely in terms of acceleration (under the aetion of forees produced, for example, by stretehed springs). One cxperiments with objects of all sorts of materials, in different states of aggregation, and so on. It then turns out that, in each and every instance, the gravitational charge is strictly proportional to an independently established quantity, the inertial mass. Is this just a remarkable coincidence, 279 Mass and weieht or does it point to something very fundamental? For a long time this apparent coincidence was regarded as one of the unexplained mysteries of nature. It took the sagacity of an Einstein to suspect that gravitation may, in a sense, be equioalenI to acceleration. Einstcin's "postulate of equivalence," that the gravitational charge q and the inertial mass m are measures of the same quantity, provided the basis of his own theory of gravitation as embodied in the general theory of relativity. We shall come back to this in Chapter 12, when we discuss noninertial frames of reference. We are quite accustomcd to exploiting the proportionality of F„ to m in our use of the equal-arm balance [Fig. 8-1 8(a)]. What we are doing is balancing the torques of two forces, but what we are actually interested in is the equality of the amounts of material. We make use of the fact that, to very high accuracy, the valuc of g is the same at the positions of both masses, and we do not need to bother about what its particular value is. Thanks to the proportionality of gravitational force to mass, we could, with an equal-arm balance and a set of Standard weights, measure out a requircd quantity of any substance equally well on the earth, the moon, or Mars. The spring balance [Fig. 8-1 8(b)], on the other hand, has a calibration that depends directly on a particular value of g. Its readings are in effect readings of force, even though we use them as a basis for mea- suring out required amounts of mass. We might find it con- venient to use a spring balance for this purpose on the moon, but if its scale were marked in kilograms, we should have to mask this out and attach a fresh calibration with the help of Standard masses. Fig. 8-18 (a) Weigh- ing with an egual-arm balance — in effect a direct comparison of masses, valid whatever the value of g. (b) Weighing with a spring balance — a measurement of the gravitational force, directly dependent on the value of g. 280 Universal gravitation Fig. 8-19 Forces acling on the bob ofa simple pendulum. Newton himself recognized that the strict proportionality of the gravitational force to the inertial mass, as evidenced by the identical local acceleration of falling objects, was a key feature in his own statement of universal gravitation as expressed in Eq. (8-8). He therefore made a series of very careful pendulum experiments to test whether a pendulum of a given length, but with a variety of objects used as the bob, always had the same period. To see how this works, consider an object of inertial mass m hung on a string (Fig. 8-19). The two forces on it (ignor- ing air resistancc) are the tension T and the gravitational force F„. The tension T is at every instant pcrpendicular to the path of the pendulum bob and so has no effect on the tangential acceleration a$. The tangential acceleration is due to the tangential com- ponent of F„. From Fig. 8-19, Fe = ma» = —F a smd from which a« = — (Fg/m)sin 9 (8-26) At every angle 6 the acceleration ag depends on the ratio F„/m. Therefore — for given initial conditions — the vehcity of the bob at every angle 6 will be determined by this ratio. So also the period for one completc round trip will depend upon the ratio (F s /m). Newton observed the periods of pcndulums with dif- ferent bobs but with equal lengths. From his observation that the periods of all such pendulums were equal within his experi- mental error, Newton concluded that F was proportional to m to better than 1 part in 1000. More recent experiments (beginning with Baron Eiitvos in Budapest in the nineteenth century) have made use of a very clever idca that permits a static measurement. It depends on recognizing that an object hanging at rest relatioe to the earth i n fact has an acceleration toward the earth's axis because, by virtue of the earth's rotation, it is traveling in a circular path of radius r = R cos \, whcrc X is the latitudc (see Fig. 8-20). This means that a net force of magnitude mu 2 r must be acting on it, where m is the inertial mass. How is this force provided? The answer is that, when a body hangs on a string ncar the earth's surface, the string, exerting a force T, is not in quite the same 281 Mass and weight Fig.8-20 Basis of the Eotvos methodfor comparing the inerlial mass and the gravitational mass of mi object that is at rest relatice to the earth and hence is being aceeler- ated toward the earth' s axis. Fig. 8-21 Principle oftlte Eoluos lorsion- balance measurement : (o) Two approximately egual masses hang from a torsion bar. (b) I f the objects do not have identical ratios of inerlial to gracitational mass, the lensions in the sus- pending strings must be in slightly different directions. In eguilib- rium, the direction of the main supporting fiber must be interme- diate between the directions ofTi and T 2 . (c) This implies the possibility of a net torque that twists the torsion bar abotit a vertical axis. direction as the gravitational force F„. And if F„ is not strictly proportional to m, the angle between T and F„ will be different for different objects. To search for any such variations, a very sensitive torsion balance is used, carrying dissimilar objects at the two ends of a horizontal bar [Fig. 8-21 (a)]. If the directions of the tensions T, and T a are different [Fig. 8-21 (b)], there will be small horizontal components [Fig. 8-21(c)] that act in op- posite directions with respect to the horizontal bar but that givc torques in the same sense. On the other hand, if the directions of T! and T 2 are identical, even if their magnitudes are not quite the same, there is no net torque tending to twist the torsion fiber. To test for the existence of any such torque, Eotvos placed the whole apparatus i n a case that could be rotated. The hori- jyi '■6 T, sin B (a) (b) r, sin i* (c) 282 Universal eravitation Fig. 8-22 To see whether a net torque exisls in the Eotvos experiment, the whole apparatus is turned through 180". This would recerse the sense ofthe torque. zontal beam carrying the two masses was aligned in an east-west direction [Fig. 8-22(a)] and a reading was taken of its position with respect to the case. The whole system was then rotated through 180°, as inFig. 8-22(b). If you analyze both situations on the basis of Fig. 8-21, you will find that, with respect to the center line of the case, the angle of twist would be reversed by this operation; hence, if any net torque existed, its existence would be revealed. More recently, some elegant modernized experiments of this type have been performed by R. H. Dicke and his collab- orators. ' By such experiments it has been shown that the strict proportionality of F B to m holds to 1 part in 10 10 or better. The description of the above experiments points to a closely related phenomenon — a systematic variation with latitude of the measured weight of an object. If we take the idealization of a perfectly spherical earth (Fig. 8-23), the equilibrium of the object is maintained by applying a force of magnitude W at an angle a to the radius such that the following conditions are satisfied: Wsina = moi 2 rsin\ F„ — Jfcosa = mui 2 rcos\ where r = R cos X. Since a is certainly a very small angle, it is justifiable to put cos « « 1 in the second equation, thus giving the result W « F g - mo> 2 R cos 2 X 'See R. H. Dicke, Sci. Am., 205, 84 (Dec. 1961). 283 Mass and weight Fig. 8-23 The force needed to balance the weiglil of an object is different 'm bolh direction and magni- tude from t he force of gravity. It follows that W(\) « Wo + mu 2 R sin 2 X where W is the measured weight on the equator. Putting iy = mgo, we can also obtain a corresponding expression for the latitude dependence of g: l?(X) = 80 + <* 2 R sin 2 X If in this expression we substitute w = 27r/86,400sec -1 and R = 6.4 X 10° m, we find w 2 /? « 3.4 X 10~ 2 m/sec 2 , which with g « 9.8 m/sec 2 gives us g(\) « 9.8(1 + 0.0035 sin 2 X) m/sec 2 This formula is more successful than it deserves to be, for we have no right to ignore the significant flattening of the earth, due again to the rotation, which makes the equatorial radius of the earth about 1 part in 300 greater than the polar radius. This eliipticity has two consequences: It puts a point on the equator farther away from the earth's center than it otherwise would bc, but it also in effect adds an extra belt of gravitating material around the equatorial region. The resultant value of g at sea level, taking these effects into account, is quite well de- scribed by the following formula: g(\) = 9.7805(1 + 0.00529 sin 2 X) (8-27) Thus our simplc calculation has the correct form, but its value for the numerical coefRcient of the latitude-dependent correction is only about two thirds of the true figure. 284 Universal gravitation WE1GHTLESSNESS It is appropriate, after thc detailed discussion of the relations among mass, gravitational force, and weight, to devote a few words to the propcrty that is called weightlessness. The very explicit distinction that we have drawn between the gravitational force on an object and its measured weight makes use of what is called an operational definition of the latter quantity. The weight, as we have defined i t, is the magnitude of the force that will hold an object at rest relative to the earth. Our definition of weightlessness derives very naturally from this: An object is weight less whenever i t is in a state of completely free fail. In this state each part of the object undergocs the same acceleration, of whatcvcr value corresponds to the strcngth of gravity at its location. (In saying this we assume that g does not change ap- preciably over the linear extent of the object.) An object that is prevented from falling, by being restrained or supported, inevitably has internal stresses and deformations in its equi- librium state. This may become very obvious, as when a drop of mercury flattens somewhat when it rests on a horizontal sur- face. Ali such stresses and deformations are removed in the weightless state of free fail. The mercury drop, for cxample, is free to take on a perfectly spherical shape. The above definition of weightlessness can be applied in any gravitational environment, and this is the way it should be. The bizarre dynamical phenomena of life in a space capsule do not depend on getting into regions far from the earth, where the gravitational forces are much reduced, but simply on the fact that the capsule, and everything in it, is falling freely with the same acceleration, which in consequence goes undetected. For ex- ample, if a spacecraft is in orbit around the earth, 200 km above the earth's surface, the gravitational force on the spacecraft, and on everything iri it, is still about 95% of what one would measure at sea level, but the phenomena associated with what we call weightlessness are just as pronounced under these con- ditions as they are in another spacecraft 200,000 miles from earth, where the earth's gravitational attraction is down to j^Vo of that at the earth's surface. In both situations an object released inside the spacecraft will remain poised in midair. The same would be true in a spacecraft that was simply falling radially toward the earth's center rather than pursuing a cireular or elliptical orbit around the earth. When viewcd in these terms the phe- 285 Weightlessness nomena of weightlessness are not in the least mysterious, al- though they are still startling because they conflict so strongly with our normal experience. LEARNING ABOUT OTHER PLANETS The recognition of the universality of gravitation gives us a powerful tool for obtaining Information about planets other than the earth, and indeed about celestial objects in general. In particular, if a planet has satellites of its own, we can find its mass by an analysis exactly similar to the one we used for de- ducing the mass of the sun from the motions of the planets themselves. This provides the simplest way of finding the mass of any planet that has satellites. Such satellites, if a planet has more than one of them, also provide a further test of Kepler's third law, taking the planet itself as the central gravitating body. Newton himself applied an analysis of this kind to Jupiter, using data for its four most prominent satellites. These were the satellites (or "moons") that made history when Galileo dis- covered them with his new astronomical telescope in 1610 (see pp. 287-290). Figure 8-24(a) shows their changing positions as seen through a modern telescope, and Fig. 8-24(b) reproduces a few of the sketches that Galileo himself made, night after night, over a period of many months. Figure 8-24(c) is a graph con- structed from Galileo's quantitative records, using the readings that can be unambiguously associated with the outermost of the four satellites. A period of about 16 days can be inferred. Galileo had no hesitation in interpreting his observations in terms of the four satellites following circular orbits that were seen edgewise — giving, as we would describe it, the appearance of simple harmonic motion at right angles to our line of sight. On the basis of further measurements Galileo arrived at rather accurate values of the orbital periods of all four satellites and at moderately good values of the orbital radii expressed as multiples of the radius of Jupiter itself. Newton, in the Principia, used similar data of greater pre- cision, obtained by his contemporary John Flamsteed. Table 8-3 (p. 290) lists these data, and in Fig. 8-25 they are plotted loga- rithmically (cf. Fig. 8-6) in such a way as to show how they give another demonstration of the correctness of Kepler's third law. The slope is accurately f. The use of Jupiter's radius as a unit for measuring the orbital radii was not merely a convenience. As we have already 286 Universal gravitation noted in our discussion of the mass of the suri, the absolute scale of the solar system was not known with any great certainty in Newton's day. It is interesting, however, that using the data as presented in Table 8-3, without absolute values of the radii, one can deduce the mean density, pj, of Jupiter. By analogy with our analysis of earth satellites [p. 266, and in particular Eq. (8-15)], we have 2xr (GMj\ 1/2 whence M j = — Putting Mj = ^ Pj Rj 3 we get i '■> _ 3t n Substituting n*/T 2 ~ 7.5 X 10 -9 sec -2 , and G - 6.67 X 10- n m 3 /kg-sec 2 , wefind pj m 1050 kg/m 3 i.e., about the same density as water. If a planet does not have satellites of its own, the magnitude of its mass may be inferable from a detailed analysis of the tiny disturbing effects — called perturbations — that planets exert on one another's orbits. This technique has been used for Mercury and Venus. The unraveling of these mutual interactions is a complicated and difficult matter, however, and in at least one case it posed a problem that was not adequately answered for a long time. This was the interaction between the two most massive planets, Jupiter and Saturn, which caused irregularities of a very puzzling kind in the orbits of both. It was even con- sidered possible that the basic law of gravitation would need to be modified slightly away from a precise inverse-square relation- ship. The solution to the mystery was finally achieved, almost a century after the publication of the Principia, by the Frcnch mathematician Laplace, building on work by his great fellow 287 Learning about othcr planels The moons of Jupiter In 1608 Hans Lippershey, in Holland, patented what may hace been the first successful lelescope. Galileo learned of this, and soon made lelescopes of his own design. His Ihird instrumen!, of more than 30 diamelers' magnification, led him lo a dramalic discouery, as recounted by him in his book, The Slarry Messenger: "On the secenth day of January in this prcsenl year 1610, al Ihe first hour of Ihe night, when I was ciewing ihe heavenly bodies with a teleseope, Jupiter presented itself lo me, and . . . I perceiced that beside Ihe planet there were three small starlets, small indeed, bui cery bright. Though I belieced them lo be among Ihe hosl offixed stars, they aroused my curiosily somewhat by appearing lo lie in an exacl straight line parallel t o the ecliplic . . . I paid no attenlion to the distances between them and Jupiter, for at Ihe oulsel l thoughl them lo be fixed stars, as l hace said. Bui returning lo the same incestigalion on January eighth — ledby what, l do not know — I found a uery different arrangemenl. . . ." Afew more night s ofobseruation were enough lo concince Galileo what he was seeing: "I hadnow [by January 11] decided beyond all question thal there exisled in Ihe heavens three stars wandering aboul Jupiter as do Venus and Mercury about the sun. . . . Nor were there just three such stars; four wanderers complele their revolutions aboul Jupiter. ... Also I measured the distances between them by means of the teleseope. . . . Moreocer I recorded the times of the observa- tions . ..for the revolutions ofthese planet s are so speedily completed that i t is usually possible t o lake eien their hourly varialions." Fig. 8~24(a) Jupiter and its four mosi prominent salellites as seen through a modern teleseope. The first and second photo- graphs illustrale Galileo's observa- tion that noticeable changes oecur within a single night. ( Yerkes Obsercatory photo- graphs) 288 Universal gravitation 4-S-A»? i '•■™y (8& 4-J«- d -7- M r. *^_^>D^ =*-*=^V* +■ ■ f-.tv-r- #r>_ -7» jf * T' .=: <a.6?M/ *■ <-* r- tf- r V J?Tb Kr. W 4- «' o v*—* «"'*•' 0***-* + Fig. 8-24(b) Facsimile of a page of Golileo's own handwritten records ofhis observations diiring the later months (July-October) of 1610. Fig. 8-24(c) A graph conslrucied from Galileo's own records, showing the periodic mol ion ofCallisto, the outermost of the four satellites visible to Mm. The period of about 16 3/4 days is clearly exhibited. M " *" '° " " * 289 The raoons of Jupiter countryman Lagrange. It turned out that a curious kind of resonance effect was at work, resulting from the fact that the periods of Jupiter and Saturn are almost in a simple arithmetic TABLE 8-3: DATA ON SATELLITES OF JUPITER- Satellile n = r/Rj Period (T) n 3 /T 2 , sec~ 2 Io Europa Ganymede Callisto 5.578 1.7699 days « 1.53 X 10 5 sec 7.4 X 10-° 8.876 3.5541 days « 3.07 X 10 5 sec 7.5 X 10~ 9 14.159 7.1650 days « 6.19 X 10 5 sec 7.5X10-° 24.903 16.7536 days « 1.45 X 10 6 sec 7.4 X IO" 9 Fig. 8-25 A log-log grapli displaying the applicabilily of Kepler's thitd law IO the Galilean satellites of Jupiter. It may be seen that Calileo's results are little dif- ferent from those ob- tained by John Flam- steed nearly 100 years later. "Thcse same data have been presented in a striking way in Eric Rogers, Physicsfor the Inauiring Mind, Princeton University Press, Princeton, N.J., 1960: r\ Satellile (milesy T 2 , (hours) 2 Io 1.803 X 10" 1.803 X 10 3 Europa 7.261 X 10 10 7.264 X 10 3 Ganymede 29.473 X 10 lfl 29.484 X 10 3 Callisto 160.440 X 10 16 160.430 X 10 3 How would you convince your friends that this close numerical coin- cidencc is not evidence of a new fundamental law? 290 Universal gravitation relationship (5Tj » 2T S )- This made large an otherwise negligible term in the perturbation, with a repetition period so long (~900 yr) that it seemed to be increasing without limit. When the mystery was finally resolved the belief in Newton's theory was, of course, strengthened still further. THE DISCOVERY OF NEPTUNE Probably the most vivid illustration of the power of the gravita- tional theory has been the prediction and discovery of planets whose very existence had not previously been suspected. It is noteworthy that' the number of known planets remained un- changed from the days of antiquity until long after Newton. Then, in 1781, William Herschel noticed the object that we now know as Uranus. He was engaged in a systematic survey of the stars, and his only clue to start with was that the object seemed slightly less pointlike than the neighboring stars. Then, having a telescope with various degrees of magnification, he confirmed that the size of the image increased with magnification, which is not true for the stars — they remain below the limit of resolution of even the biggest telescopes, and always produce images in- distinguishable from those due to ideal point sources. Once his attention had been drawn to the object, Herschel returned to it night after night and confirmed that it was moving with respect to the other stars. Also, as has happened in various other cases, it was found that the existence of the object had in fact been recorded i n earlier star maps (first by John Flamsteed in 1690). These old data suddenly became extremely valuable, because they were a ready-made record of the object's move- ments dating back through nearly a century. When combined with new measurements carried out over many months, they showed that the object (finally to be called Uranus) was indeed a member of our solar system, following an almost circular orbit with a mean radius of 19.2 AU and a period of 84 years. This is where our main story begins. Once it was discovered, Uranus and its motions became the subject of a continuing study, and cvidence began to accumulate that there were some ex- tremely small irregularities in its motion that could not be aseribed to perturbing effects from any known source. Figure 8-26(a), a tribute to the wonderful precision of astronomical observation, shows the anomaly as a funetion of time since 1690. The suspicion began to grow that perhaps there was yet another 291 The discovery of Neptune Fig. 8-26 (a) Unexplained residual devialions in the observed positions of Uranus between 1690 and 1840. (b) Basis ofascribing the devialions lo the influence of an extra planet. The arrows indicate the relalive mag- nitude of the perturbing force a t dijferent times. planet beyond Uranus, unknown in mass, period, or distance. Two men — J. C. Adams in England and U. J. LeVerrier in Francc — indepcndcntly workcd on the problem. Both men used 292 Universal gravitation as a starting point the assumption that the orbital radius of the unknown planet was almost exactly twice that of Uranus. The basis of this was a curious empirical relation, known as Bode's law (actually discovered by J. D. Titius in 1772, but publicized by J. Bode), which expresses the fact that the orbital radii of the known planets can be roughly fitted by the following formula: R„ (AU) = 0.4 + (0.3X2") where n is an appropriate integer for each planet. Putting n = 0, 1, 2, we get the approximate radii for Venus, earth, and Mars (Mercury requires n = — cc, which is hard to defend). Using n = 4, 5, and 6 one gets quite good values for Jupiter, Saturn, and Uranus. (The missing integer, n = 3, corresponds to the asteroid belt.) Figure 8-27 shows this relation of orbital radii with the help of a semilog plot; it is clear that a simple exponential relation (linear on this graph) works almost as well, but if one accepts Bode's law, then n = 7 gives r = 38.8 AU, and this is what Adams and LeVerrier used. Fig. 8-27 Graph for predicting the orbital radius of the new planet with the help of Bode's law. 293 The discovery of Neplune Given the radius, the period is automatically defined by Kepler's third law, and then it becomes possible to construct a definite picture, as shown in Fig. 8-26(b), of the way in which the new planet could alternately accelerate and retard Uranus in its orbital motion, depending on their relative positions. With the help of laborious analysis, onc can then deduce where in its orbit the new planet should be on a particular date. Adams supplied such information in October 1845 to the British Astron- omer Royal, G. B. Airy, who acknowledged Adams' letter, raised a question of detail, but otherwise did nothing. LeVerrier did not complete his own calculations until August 1846, but the astronomer to whom he wrote (J. G. Galle, in Germany) made an immediate search and identified the new planet (Neptune) on his very first night of observation. It was only about a degree from the predicted position (see Fig. 8-28). The next night it had visibly shifted, thereby confirming its planetary status. Although the discovery of Neptune is in some respects a great success story, it is also a story of luck, both good and bad, and of human frailty. Adams was really first in the field, but he received no support from his seniors (he was fresh from his bachelor's degree when he bcgan his calculations). Airy missed Fig. 8-28 Star map showing the discovery of Neptune, September 23, 1846. (From Herbert Hal/ Turner, Astronomical Dis- covery, Edward Arnold, London, 1904.) • -. v'-- O . « *. ' • f -, -■ . ■m m » a ■ i • • 4 l t r • • (•••«« • 294 Universal gravitation the credit, which he might readily have won, of being the man who first identified Neptune. But the locations that both Adams and LeVerrier predicted might well have been hopclessly mis- leading, for in their reliance on Bode's law they used an orbital radius (and hence a period) that was far from correct. The true value is about 30 AU instead of nearly 40 as they had assumed, which means that they overestimated the period by nearly 50%. ' It was therefore largely a lucky accident that the planet was so near to its predicted position on the particular date that Galle sought and found it. But let this not be taken as disparagement. A great discovery was made, with the help of the laws of motion and the gravitational foree law, and it remains as the most triumphant confirmation of the dynamical model of the universe that Newton invented. 2 The discovery of Pluto by C. Tombaugh in 1930, on the basis of a detailed record of the irregularities of Neptune's own motion, provided an echo of the original achieve- ment. GRAVITATION OUTSIDE THE SOLAR SYSTEM When Newton wrote his System of the World, nothing was known about the distances or possible motions of the stars. They simply provided a seemingly fixed background against which the dynamics of the solar system proceeded. There were ex- ceptions. A few prominent stars— e.g., Sirius— known since antiquity by naked-eye astronomy, were found to have shifted position within historic time. But the serious and systematic investigation of stellar motions was begun by William Herschel. His observations, continued and refined by his son John Her- schel and by other astronomers, revealed two classes of results. The first was the continuing apparent displacement of individual stars in a way that suggested that the solar system is itself in- volvcd in a general movement of the stars in our neighborhood, at a speed of the order of 10 miles/sec (comparable to the earth's own orbital speed around the sun). This, as it stood, was just an empirical faet. But the second type of result pointed directly ■This also means that they overestimated the mass necessary to produce the observed perturbations of Uranus. LeVerrier gave a figure of about 35 times the mass of the earth; the currently accepted value is about half this. 2 For a detailed account of the whole matler, see H. H. Turner, Astronomku! Discovery, Edward Arnold, London, 1904. A shorter but more readily accessible account may be found in an essay entitled "John Couch Adams and the Discovery of Neptune," by Sir H. Spencer Jones, in The World of Mathemaiks (J. R. Newman, ed.), Simon and Schuster, New York, 1956. 295 Gravitation outside the solar system Scale (secondsof arc) Fig. 8-29 Varialion willi lime oflhe rela- live position vector of the members ofa double-star syslem. (After Arthur Berry, A Short History of Astronomy, 1898; reprinted by Dover Publications, New York, 1961.) '1868 1864 /*" 1880 1821 S: 1885 1781 1897 •l84o" * 1895 1827 • 1890 1830 to the operation of Newton's dynamics. For the Herschels discovered numerous pairs of stars that were evidently orbiting around one another as binary systems. Figure 8-29 shows one of the best documented early examplcs, and the first to be sub- jccted to a detailed analysis in terms of Kepler's laws. (It is £-Ursae, in one of the hind paws of the constellation known as the Grcat Bear.) The period of a binary star depends on the total mass of the system, not on the individual masses. This is easily proved in the case in which the orbits are assumed to be circles around the common center of mass [see Fig. 8-30(a)]. ' The individual stars are always at opposite ends of a straight line passing through the center, C. If we write the statement of F = ma for one of the stars, say mi, we have , /W2WI1 (r. + r 2 )2 mi»t = miu r\ where w (= 2tt/T) is the angular velocity common to both stars. Hence 2 CO = Gnt2 ri(ri + r 2 ) 2 ■For a full discussion of the concept of center of mass, sce Ch. 9, p. 337. 296 Universal gravitation (a) Fig. 8-30 (a) Motion ofthe members of a binary star system with respect to the center of mass. C, for the case ofcircular orbits. (b) Direct visual euidence of the motion of a binary system — Krueger 60, photographed by E. E. Barnard. (Yerkes Obseroatory photo- graph.) (b) However, by the definition of the center of mass, we have 'i = mi + rri2 where r = n + ri It follows at once that 2 G(m\ + m 2 ) 03 = r, Thus if the distance r between the stars can be obtained by direct astronomical observation (e.g., starting from a knowledge of their angular scparation), the sum of the masses is at once determined. Finding the individual masses entails the somewhat harder job of measuring the motion of each star in absolute terms against the background ofthe "fixed" stars. Figure 8-30(b) shows convincing direct evidence of the orbital motion of an actual binary system. 297 Gravitation outside the solar system Fig. 8-31 Rotating galaxy (spiral galaxy NGC 5194 in the constellation Canes Venatici). (Photo- graph from the Hale Obsercatories.) With the development of modern astronomy, the systematic motions of our sun and its neighbors came to be seen as part of a greater scheme of movements controlled by gravity. AU around us throughout the universe were the immense stellar systems— the galaxies— most of them vividly suggesting a state of general rotation, as i n Fig. 8-31, for example. The most difficult structurc to elucidate was the onc in which we ourselves are embedded, i.e., the Milky Way galaxy. It finally became clear, however, that its basic structure is very much like that of Fig. 8-31, and that in it our sun is describing some kind of orbit around the center, with a radius of about 3 X 10 20 m (= 30,000 light-years) and an estimated period of about 250 million years (=» 8 X 10 ,5 sec). Using these figures we can infer the ap- 298 Universal gravitation proximate gravitating mass, inside the orbit, that would define this motion. From Eq. (8-24) we have .23 With G = 7 X 10~" m 3 /kg-sec 2 , we find that m ~ 40 v 3 X1 ° G1 « 3 x 10 41 ke Since the mass of the sun (a typical star) is about 2 X 10 30 kg, we see that a core of about 10" stars is implied. This is not really a figure that can be independently checked. It is a kind of ultimate tribute to our belief in the universality of the gravita- tional law that it is confidently used to draw conclusions like those above concerning masses of galactic systems. EINSTEIN'S THEORY OF GRAVITATION We have described carlier how Newton rccognized that the proportionality of weight to inertial mass is a fact of fundamental significance; it played a central role in leading him to the con- clusion that his law of gravitation must be a general law of nature. For Newton this was a strictly dynamical result, ex- pressing the basic properties of the force law. But Albert Einstein, in 1915, looked at the situation through new eyes. For him the fact that all objects fail toward the earth with the same accelera- tion g, whatcver their size or physical state or composition, implied that this must be in some truly profound way a kinematic or geometrical result, not a dynamical one. He regarded it as being on a par with Galileo's law of inertia, which expressed the tendeney of objects to persist in straight-line motion. Building on these ideas, Einstein developed the theory that a planet (for example) follows its charaeteristie path around the sun because in so doing it is traveling along what is called a geodesic line — that is to say, the most economical way of getting from one point to another. His proposition was that although in the absence of massive objects the geodesic path is a straight line in the Euclidean sense, the presence of an extremely massive object such as the sun modifies the geometry locally so that the geodesies become curved lines. The state of affairs in the vicinity 299 Einsteirrs theorv o! gravitation of a massive object is, in this view, to be interpreted not in terms of a gravitational field of force but in terms of a "curvature of space" — a facile phrase that covers an abstract and mathe- matically complex description of non-Euclidean geometries. For the most part the Einstein theory of gravitation gives results indistinguishable from Ncwton's; the grounds for pre- ferring it might seem to be conceptual rather than practical. But in one celebrated instance of planetary motions there is a real discrepancy that favors Einstein's theory. This is in the so-called "precession of the perihelion" of Mercury. The phe- nomenon is that the orbit of Mercury, which is distinctly el- liptical in shapc, very gradually rotates or precesses in its own planc, so that the major axis is along a slightly different direction at the end of each complete revolution. Most of this precession (amounting to about 10 minutes of arc per century) can be understood in terms of the disturbing effects of the other planets according to Newton's law of gravitation. ' But there remains a tiny, obstinate residual rotation equal to 43 seconds of arc per century. The attempts to explain it on Newtonian theory — for example by postulating an unobserved planet inside Mercury's own orbit — all came to grief by conflicting with other facts of observation concerning the solar system. Einstein's theory, on the other hand, without the use of any adjustable parameters, led to a calculated precession rate that agreed exactly with observation. It corresponded, in effect, to the existence of a very small force with a different dependence on distance than the dominant l/r 2 force of Newton's theory. The way in which a disturbing effect of this kind causes the orbit to precess is dis- cussed in Chapter 13. Other cmpirical modifications of the basic law of gravitation — small departures from the inverse- square law — had been tried before Einstein developed his theory, but apart from their arbitrary character they also led to false predictions for the other planets. In Einstein's theory, however, it emerged automatically that the size of the disturbing term was proportional to the square of the angular velocity of the planet and hence was much more important for Mercury, with its short period, than for any of the other planets. ■The apparent amount of precession as viewed from the earth is aclually about 1.5° per century, but most of this is due to the continuous change in the direction of the earth's own axis (the precession of the equinoxes — see Chapter 14). 300 Universal gravitation PROBLEMS 8-1 Given a knowledge of Kepler's third law as it applies to the solar system, together with the knowledge that the disk of the sun subtends an angle of about £° at the earth, deduce the period of a hypothetical planet in a circular orbit that skims the surface of the sun. 8-2 It is well known that the gap between the four inner planets and the five outer planets is occupied by the asteroid belt instead of by a tenth planet. This asteroid belt extends over a range of orbital radii from about 2.5 to 3.0 AU. Calculate the corresponding range of periods, expressed as multiples of the earth's year. 8-3 It is proposed to put up an earth satellite in a circular orbit with a period of 2 hr. (a) How high above the earth's surface would it have to be? (b) If its orbit were in the plane of the earth's equator and in the same direction as the earth's rotation, for how long would it be continuously visible from a given place on the equator at sea level? 8-4 A satellite is to be placed in synchronous circular orbit around the planet Jupiter to study the famous "red spot" in Jupiter's lower atmosphere. How high above the surface of Jupiter will the satellite be? The rotation period of Jupiter is 9.9 hr, its mass Mj is about 320 times the earth's mass, and its radius Rj is about 11 times that of the earth. You may find it convenient to calculate first the gravita- tional acceleration gj at Jupiter's surface as a multiple of g, using the above values of Mj and R Jt and then use a relationship analogous to that developed in the text for earth satellites [Eq. (8-16) or (8-17)]. 8-5 A satellite is to be placed in a circular orbit 10 km above the surface of the moon. What must be its orbital speed and what is the period of revolution ? 8-6 A satellite is to be placed in synchronous circular earth orbit. The satellite's power supply is expected to last 10 years. If the maxi- mum acceptable eastward or westward drift in the longitude of the satellite during its lifetime is 10°, what is the margin of error in the radius of its orbit? 8-7 The springs found in retractable ballpoint pens have a relaxed length of about 3 cm and a spring constant of perhaps 0.05 N/mm. Imagine that two lead spheres, each of 10,000 kg, are placed on a frictionless surface so that onc of thesc springs just fits, in its un- compressed state, between their nearcst points. (a) How much would the spring be compressed by the mutual gravitational attraction of the two spheres? The density of lead is about 1 1,000 kg/m 3 . 301 Problcms (b) Let the system be rotatcd in the horizontal plane. At what frequency of rotation would the presencc of the spring become ir- relevant to the separation of the masses? 8-8 During the cighteenth century, an ingenious attempt to find the mass of the earth was made by the British Astronomer Royal, Nevil Maskelyne. He observed the extent to which a plumbline was pulled out of true by the gravitational attraction of a mountain. The figure illustrates the principle of the method. The change of direction of the plumbline was measured between the two sides of the mountain. (This was donc by sighting on stars.) After allowance had been made for the change in direction of the local vertical because of the curvature of the earth, the residual angular difference a was given by 2Fm/Fb, where ±Fm is the horizontal force on the plumb-bob due to the moun- tain, and Fe = GM E m/RE^. (m is the mass of the plumb-bob.) The value of a is about 10 seconds of arc for measurements on opposite sides of the base of a mountain about 2000 m high. Suppose that the mountain can be approximated by a cone of rock (of density 2.5 times that of water) whose radius at the base is equal to the height and whose mass can be considered to be concentrated at the center of the base. Deduce an approximate value of the earth's mass from these figures. (The true answer is about 6 X 10 24 kg.) Compare the gravitational deflection a to the change of direction associated with the earth's curvature in this experiment. 8-9 Imagine that in a certain region of the ocean floor there is a roughly cone-shaped mound of granite 250 m high and 1 km in diam- eter. The surrounding floor is relatively flat for tens of kilometers in all directions. The ocean depth in the region is 5 km and the density of the granite is 3000 kg/m 3 . Could the mound's prcscnce be de- tected by a surface vessel equipped with a gravity meter that can detect a change in g of 0.1 mgal? (Hint: Assume that the field produced by the mound at the surface can be approximated by the field of a mass point of the same total mass locatcd at the level of the surrounding floor. Note that in calculating the change in g you must keep in mind that the mound has displaccd its own volume of water. The density of water, even at such depths, can be taken as about equal to its surface value of about 1000 kg/m 3 . 302 Universal gravitation 8-10 Show that the period of a particle that moves in a circular orbit close to the surface of a sphere depends only on G and the mean density of the sphere. Deduce what this period would be for any sphere having a mean density equal to the density of water. (Jupiter almost corresponds to this case.) 8-11 Calculate the mean density of the sun, given a knowledge of G, the length of the earth's year, and the fact that the sun's diameter subtends an angle of about 0.55° at the earth. 8-12 An astronaut who can lift 50 kg on earth is exploring a planetoid (roughly spherical) of 10 km diameter and density 3500 kg/m 3 . (a) How large a rock can he pick up from the planetoid's surface, assuming that he finds a well-placed handle? (b) The astronaut observes a rock falling from a cliff. The rock's radius is only 1 m and as it approaches the surface its velocity is 1 m/sec. Should he try to catch it? (This is obviously a fanciful problem. One would not expect a planetoid to have cliffs or loose rocks, even if an astronaut were to get there in the first place.) 8-13 It is pointed out in the text that a person can properly be termed "weightless" when he is in a satellite circling the earth. The moon is a satellite, yet it is noted in many discussions that we would weigh | of our normal weight there. Is there a contradiction here? 8-14 A dedicated scientist performs the following experiment. After installing a huge spring at the bottom of a 20-story-high elevator shaft, he takes the elevator to the top, positions himself on a bathroom scale inside the airtight car with a stopwatch and with pad and pencil to record the scale reading, and directs an assistant to cut the car's support cable at I = 0. Presuming that the scientist survives the first encounter with the spring, sketch a graph of his measured weight versus time from / = up to the beginning of the second bounce. (Note: Twenty stories is ample distance for the elevator to acquire terminal velocity.) 8-15 A planet of mass M and a single satellite of mass M/10 revolve in circular orbits about their stationary center of mass, being held together by their gravitational attraction. The distance between their centers is D. (a) What is the period of this orbital motion? (b) What fraction of the total kinetic energy rcsides in the satellites? (Ignore any spin of planet and satellite about their own axes.) 8-16 We have considered the problem of the moon's orbit around the earth as if the earth's center represented a fixed point about which the motion takes place. In fact, however, the earth and the moon revolve about their common center of mass. 303 Problems (a) Calculate the position of the center of mass, given that the earth's mass is 81 limes that of the moon and that the distance be- tween their centers is 60 earth radii. (b) How much longer would the month be if the moon were of negligible mass compared to the earth ? 8-17 The sun appears to be moving at a speed of about 250km/sec in a circular orbit of radius about 25,000 light-years around the center of our galaxy. (One light-year^ 10 10 m.) The earth takes 1 year to describe an almost circular orbit of radius about 1.5 X 10" m around the sun. What do these facts imply about the total mass responsible for keeping the sun in its orbit? Obtain this mass as a multiple of the sun's mass M. (Note that you do not need to introduce the numerical value of G to obtain the answer.) 8—18 (A good problem for discussion.) In 1747 Georges Louis Lesage explained the inverse-square law of gravitation by postulating that vast numbers of invisible particles were flying through space in all directions at high speeds. Objects like the sun and planets block these particles, leading to a shadowing effect that has the same quantitative result as a gravitational attraction. Consider the arguments for and against this theory. (Suggestion: First consider a theory in which opaque objects block the particles completely. This proposal is fairly easy to refute. Next consider a theory in which the attenuation of the particles by objecls is incomplete or even vcry small. This theory is much harder to dismiss.) 8-19 The continuous output of energy by the sun corresponds (through Einstein's relation E = Mc 2 ) to a steady decrease in its mass M, at the rate of about 4 X 10° tons/sec. This implies a progressive in- crease in the orbital periods of the planets, because for an orbit of a given radius we have T~ A/ - " 2 [cf Eq. (8-23)]. A precise analysis of the effect must take into account the fact that as M decreases the orbital radius itself increases — the planets gradually spiral away from the sun. However, one can get an order- of-magnitude estimate of the size of the effect, albeit a little bit on the low side, by assuming that r remains constant. (See Problem 13-21 for a more rigorous treatment.) Using the simplifying assumption of constant r, estimate the approximate increasc in the length of the year resulting from the sun's decrease in mass over the timc span of accurate astronomical ob- servations— about 2500 years. 8-20 It is mentioned at the end of the chapter how Einstein's theory of gravitation leads to a small correetion term on top of the basic Newtonian foree of gravitation. For a planet of mass m, traveling at speed o in a circular orbit of radius r, the gravitational foree becomes 304 Universal gravitation in effect the following: _ GMm r2 (•♦«a where c is the speed of light. (Correction terms of the order of v 2 /c 2 are typical of relativistic effects.) (a) Show that, if the period under a pure Newtonian force GMm/r 2 is denoted by 7"o, the modified period T is given approxi- mately by -»(«-3») (Treat the relativistic correction as representing, in effect, a small fractional increase in the value of G, and use the value of v corre- sponding to the Newtonian orbit.) (b) Hence show that, in each revolution, a planet in a circular orbit would travel through an angle greater by about 24T 3 r 2 /c 2 To 2 than under the pure Newtonian force, and that this is also expressible as 6irGM/c 2 r, where M is the mass of the sun. (c) Apply these results to the planet Mercury and verify that the accumulated advance in angle amounts to about 43 seconds of arc per century. This corresponds to what is called the precession of its orbit. 305 Problcms God . . . created matter with motion and rest in its parts, and . . . now conserves in the universe, by His ordinary operations, as much of motion and of rest as He originally created. rene descartes, Principia Philosophiae (1644) 9 Collisions and conservation laws in this chapter we shall be discussing some concepts and results that are at the very heart of mechanics. They are rooted in the fact that it takes at least two particles to make a dynamical system. Until now we have rather glossed over that fact, by talking in terms of individual particles subjected to forces of various kinds. Thus, for example, the motion of a planet around the sun, or of an electron between the deflection plates of a cathode-ray tube, was discussed as the problem of a single par- ticle exposed to the force supplied by some completely immovable body or structure. But this is a very special way of looking at things and is not in general justified. The sun pulls on a planet and gives it an acceleration, to be sure, but the law of gravitation doesn't play favorites — it has a completely symmetrical form — and the planet pulls back on the sun with an equal and opposite force. So the sun must accelerate, too, and will follow some kind of path under the combined action of all the planets. It happens that the sun is hundreds of times more massive than the rest of the solar system put together, so that to a first approximation we can ignore its wanderings. But this is only an accident of disparity. The basic dynamical system is made up of two inter- acting particles, the motions of both of which must be con- sidered. The experimental study of such systems, via collision processes, was in fact the true starting point of dynamics. The study of collisions has lost none of its importance to physics in 307 the 300 years since it first became a subject of exact investigation ; let us then consider it with care. THE LAWS OF IMPACT In 1668 the Royal Society of London 1 put out a request for the experimental clarification of collision phenomena. Contribu- tions were submitted shortly afterward by John Wallis (mathe- matician), Sir Christopher Wren (architect), and Christian Huygens (physicist, and Newton's great Dutch contemporary). The results embody what we are familiar with today as the rules governing exchanges of momentum and energy in the collision between two objects. Most importantly, they involve at one stroke the concept of inertial mass and the principle of con- servation of linear momentum. These experiments revealed that the decisive quantity in a collision is what Newton called "the quantity of motion," defined as the product of the velocity of a body and the quantity of matter in it— this latter being what we have called the inertial mass. We have already (in Chapter 6) discussed Newton's straight- forward concept of mass: how he was quite ready to assume that the total mass of two objects together is just the sum of their separate masses, and how he took it for granted that the masses of difierently sized portions of a homogeneous material are pro- portional to their volumes. You will recall from our earlier diseussion that these assumptions, however reasonable they may seem, are not always strictly justified. On the other hand, their use does not lead to detectable inconsistencies in the dynamics of ordinary objects. And what Newton and his contemporaries found was that, if these commonsense ideas were accepted as a quantitative basis for relating masses, then a very simple de- seription of all collisions could be made, which expressed mathe- matically is as follows: m\u\ + m-iu-i = mwi + /M2t"2 (9-1) wherc M] and u t are the velocities before impaet, and v x and v 2 are the velocities after impaet (a one-dimensional motion is assumed). This is a very powerful generalization, because it 'The Society was formally chartered in 1662. (Newton was its President from 1703 until his death in 1727.) Several other great European seientific academies came into existence at about the same period. 308 Collisions and conservalion Uiws applies to quite different sorts of collisions— those with almost perfect rebound, such as occur between two glass spheres, down to thosc with no rebound at all, as between two lumps of putty. Qualitatively, collisions can be described in terms of their degree of elasticity— i.e., bounciness. A quantitative but purely em- pirical measure of this, used by Newton and others, is the ratio of v 2 — v y (the relative velocity after impact) to Ui — u> (the relative velocity before impact). If this ratio is zero, the collision is called completely inelastic. If the ratio is unity, the collision is called elastic (somctimes, more vividly, "perfectly elastic"). A hardened steel ball is much more elastic than a rubber ball in this sense. In the early investigations Wallis confined his studies to completely inelastic collisions, Wren and Huygens to almost perfectly elastic objects, and Newton, some time later, added experiments on objects of intermediate elasticity. THE CONSERVATION OF LINEAR MOMENTUM The physicist is always on the lookout for quantities that remain conserued (i.e., unehanged) in physical processes. Once he has discovered such quantities, they become powerful tools in his analysis of phenomena. They start out by being just an aid in codifying pasi experience. But as they are found to be applicable to more and more new instances, their value grows and one begins to make confident predielions with their help. The statement about conservation of a particular quantity is promoted to the status of a conservation law. If in some new instance the con- servation law appears to break down, one's faith in the law may be so great that one hunts around for the missing piece. Should it be found, the conservation law is strengthened still further. Thus, for example, the law of conservation of mass in chemical reaetions is accepted as a guide to all possible measurements on the masses of the reaeting materials. When the chemical balance was first applied to the study of chemical reaetions, it appeared that in some processes mass was gained, in some it was Iost, and in some it remained unehanged. But when Lavoisier proved, through numerous examples, that mass was merely transferred, and that in a elosed system it was conserved, the whole seheme became clear. Chemists could then exp]oit the conservation law. For example, they could confidently infer the mass of a gaseous produet (eseaping from an unclosed system) from easily made measurements on solid and Iiquid reactants. 309 The conservation of linear momentum Some of the most powerful aspects of our physical descrip- tion of the world are embodied in statements about conserved quantities. In mechanics the law of conservation of linear mo- mentum is such a statement — one might even claim that it is the most important single principle in dynamics. It is based directly on the results of the experiments on impact as sum- marized in Eq. (9-1). If, for a given particle, we introduce the single word momentum to describe the product mv, then we have a compact statement: Tbe total momentum of the system of two colliding particles re- mains unchanged by the collision, i. e., the total linear momentum is a conserved quantity. Underlying this generalization concerning two-particle col- lisions is the tacit assumption that the system is effectively isolated — the particles interact with each other but not with anything else. In the experiments of Newton and others this was achieved by hanging the colliding objects on long strings, so that they swung as pendulums and collided at the lowest point of their swing. In the brief duration of the impact, therefore, the objects were essentially free of all horizontal forces except those provided by their mutual interaction. Since the momentum carried by an object is given, in Newton's words, by "the velocity and the quantity of matter conjointly," — i.e., by the value of the product mv and not by the values of m and v separately 1 — it is convenient to introduce a single symbol, p, to represent the momentum of a mass m moving along a given line with velocity v. The relation Pi + P'i = const. (9-2) then describes the conservation of momentum in two-body collisions of the type studied by Newton and his contemporaries. MOMENTUM AS A VECTOR QUANTITY We have based our discussion so far on one-dimensional col- lisions. It is important to appreciate, however, that — as is apparent from its definition — the momentum of a particle is a vector quantity having the same direction as the velocity of the ■So that, for example, a body of mass \m traveling with velocity 2v has the same momentum as m traveling with v. 310 Collisions and conservation laws particle. Thus our statement of the conservation of linear mo- mentum in a collision between two bodies should really be written as follows: Pli + P2i = pi/ + P2/ (9-3) where the subscripts i and / are used to denote initial and final values, respectively (i.e., precollision and postcollision). This single vector equation defines the magnitude and the direction of any one of the momentum vectors if the other three are known. It will very often be convenient to separate Eq. (9-3) into three equations in terms of the resolved parts of the vectors along three mutually orthogonal axes (x, y, z). Each of these component equations must then be separately satisfied. Thus, for example, if two bodies, of masses m i and m 2 , have initial and final velocities U|, u 2 and Vi, v 2 , Eq. (9-3) becomes «mi + m2U2 = mivi + /H2V2 (9-4) which contains the following three independen! statements: m\u\ x + nt2U2r = fltiPia + m 2 D 2 z miui,, + nt2U 2 y = miviy + m 2 02 V (9-5) ntiui, + JW2«2. = mivi, + /W202. In carrying out actual numerical calculations this resolution of the vectors will often be necessary. But manipulations involving unspecified masses and velocities are best made in terms of the unresolved equations (9-3) or (9-4), without reference to any particular coordinate system. This is, indeed, one of the main strengths (and economies) of using vector notation. Example. An object of mass 5 kg, traveling horizontally on a frictionless surface at 16 m/sec, strikes a stationary object of mass 3 kg. After the collision, the 5-kg object is observed to have a velocity of magnitude 1 2 m/sec at 30° to its original direction of motion, as shown in Fig. 9-1. What is the velocity of the 3-kg object? We can choose an xy plane that contains all the velocity vectors. Let the +x axis lie along the original direction of motion of the 5-kg object (particle 1). Let us denote the initial momentum of the 5-kg object as p u, and the final momenta as pi/ and p 2 /. Then we have Pii : = Pi/ + P2/ (sincep 2 , = 0) P2/ = Pl. - Pl/ = Pli + ( — Pl/) 311 Momentum as a vector quantity Fig. 9-1 Conserva- | t ion ofthe total vector | momentum 'm a sim- ple collision. Figure 9-1 shows the vector construction by which p 2 / can be found. The length of p h represcnts, on some appropriate scale, the initial momentum of 80 kg-m/sec. The length of — Pi/ is 60 kg-m/sec, and its direction is as shown. The length and direction of p 2 / can be found directly from the vector triangle, and v 2 / is found as p 2 /Aw 2 . Alternatively, we can write down the momentum conserva- tion in component form. First, let us list the known quantities: m\ = 5 kg W2 = 3 kg u\ z = 16 m/sec «2x = «1„ = K2„ = v\ x = 6\/3 m/sec i;i„ = 6 m/sec Thus we have [using Eq. (9-5)] Along x: 80 = 30\/3 + 3t> 2 , Along^: = 30 + 3l>2» Hence U2l = (80 - 30\/3)/3 « 9.3 m/sec l- 2 „ = — 10.0 m/sec f 2 = [(9.3) 2 + (10.0) 2 ]" 2 « 13.6 m/sec The direction of v 2 is at an angle d to the x axis such that „ V2 U 10.0 tan 5 = -S = - — :- «2x 9.3 9 « -47° 312 Collisions and conservation laws Notice that this result has been obtained through momentum conservation alone; it requires no knowledge of the detailed interaction between the objects. Relating the momentum changes of the individual objects to the forces acting on them during the collision will, however, be our next concern. ACTION, REACTION, AND IMPULSE Isolation boundary No. 1 We shall begin this discussion with an analysis along the lines that Newton himself used. Later we shall draw attention to a somewhat different approach, which can be more readily adapted to relativistic dynamics, although it is in harmony with Ncwton's analysis within the confines of classical mechanics. Newton interpreted the collision experiments from the stand- point that the concept of inertial mass of an individual object is already established by experiments and arguments like those we made use of in Chapter 6. In that case, Eq. (9-4) is a summary of the actual experimental observations; the *'quantity of motion" is conserved. One can then, by using F = ma, draw conclusions about the forces acting during the collision. A collision is a process involving two objects, each of which exerts a force on the other. Object 1 exerts a force Fi 2 on object 2; object 2 exerts a force F 2 i on object 1 (Fig. 9-2). We make no assumptions about the relationship between the two forces, except that they act for equal times. This last assumption is certainly reasonable, because we recognize that the forces come into being as a result of the collision, and surely the duration of the collision must be the same for both objects. ' We can then take the statement of F = ma as it applies to each object separately: F21 = miai F12 = »1282 l.e., F21 F12 ai = &2 = Fig. 9-2 Inferring the equalily of action and reaction forces from the fact of momentum conservation. (9-6) ■Yet this is anolher of those intuitively "obvious" conditions that is not bind- ing and indeed has to bc qualified in some of the collision problems that re- quire the use of special relativity. 313 Action, reaction, and impulse Suppose, to simplify this present argument, that each force re- mains constant throughout a collision that lasts for a time At. Then we have Tl = ui + — A/ v 2 - u 2 + — At (9-7) mi rri2 where Ui and a 2 are the initial velocities of the two objects and ?i and v 2 are their final velocities. From these equations we therefore have mivi = /miui + F2jA/ W2V2 = W2U2 + F\2At Adding these two, we thus get /KlVl + WJ2V2 = /MlUl + /M2U2 + (F21 + Fi2>A/ (9-8) Experimentally, however, we have Eq. (9-4). We deduce, there- fore, that F21 + F J2 = i.e., F21 = -F12 (9-9) This is, of course, the famous statement known usually as Newton's third law, that "action and reaction are equal and opposite." We have already made extensive use of this result, tacitly or explicitly, earlier in this book, and in Chapter 4 we indicated the kind of experimental support that one can supply for it in static equilibrium situations. Newton, in developing the result for dynamic situations, with forces changing from instant to instant, made it quite clear that he regarded Eq. (9-9) as an in- ference from observattons. In the Principia he describes an experiment in which he floated a magnet and a piece of iron on water, and released them from rest. He noted that there was no motion of the combined mass after the magnetic attraction had pulled them together, and took this as a demonstration of what he himself called the third law of motion. His description of his pendulum collision cxperiments, also in the Principia, is couched in the same terms. Kaving introduced these forces of interaction, we can now relate them to the changes of momentum of the individual 314 Collisions and conservation laws objects in a collision. Thus, if At is the duration of the collision and F 2 i is a constant force exerted by particle 2 on particle 1, the change of momentum Api of particle 1 in the collision is given by F21A/ = Api More generally, if any constant force F acts on a particle for some short time interval A/, the change of momentum that it generates is given by FA/ = Ap As we noted in Chapter 6, Newton's own approach to dynamics was rooted in this equation rather than in F = ma. We also introduced the terminology by which the product F At is called the impulse of the force in A/. If F is varying in magnitude and/or direction during the time span At, one can proceed to the limit of vanishingly small time and so obtain the following equation: (Newton's law reformulated) F = — (9-10) at This eguation, written on the assumption that F is the net force acting on a particle, then becomes the basic statement of Newtonian dynamics. It is, in a sense, broader in scope than F = ma, or at least a more emcient statement of it. For example, a given force applied in turn to a number of different masses causes the same rate of change of momentum in each but not the same acceleration. In short, we come to recognize that momentum is a valuable single quantity to be accepted in its own right and most importantly for its property of being conserued in an isolated system of interacting particles. In our earlier discussion of the collision problem, we took the forces of interaction as being constant in time. In most cases that would be quite unrealistic. It is easy to see, however, that through the integration of Eq. (9-10) we obtain the net mo- mentum change caused by a varying force. Thus, for example, if a force has some arbitrary variation between / = and / = At, the momentum change that it generates is given by Ap = / Fdt (9-11) Jo In a two-body collision, the duration of which is Ar, all that 315 Action, rcaction. and impulse we actually observe is that the total momentum after the collision is equal to the total momentum before the collision. In terms of the impulses Api and Ap 2 given to the separate objects this result is expressed by the condition Api + Ap 2 = From this we infer that / F 2 i dt = - / Pu dt In principle, Fi a and F 2 i could have quite unrelated values at any particular instant, as long as the above integrals are equal. However, failing any evidence to the contrary we assume that they are equal and opposite at each and every instant. Thus in a one-dimensional collision the graphs of these forces as a func- tion of time, whatever their exact form, are taken to be mirror images of each other as shown in Fig. 9-3. It is important to realize, however, that this is a postulate. And it is not always true! There is no difficulty as far as "contact" collisions between ordinary objects are concerned. But in situations in which objects influence one another at a distance, as for example through the long-range forces of electricity or gravitation, Newton's third law may cease to apply. For no interaction is transmitted in- stantaneously, and if the propagation time cannot be ignored in comparison with the time scale of the motion, the concept of instantaneous action and reaction can no longer be used. A simple mechanical model of such a delayed interaction is sug- Fig. 9-3 Corresponding variatioiis of action and reaction forces during the course of a collision. Fn O O F„ A/ A/ 316 Collisions and conservation laws Fig. 9-4 Interaction wilh a time delay, mediated by particles trauersing the gap between two separated objects. gested in Fig. 9-4. A cart, A, carries a gun that fires off a stream of bullets with speed V. At a distance L away there is a second cart, B, carrying a block in which the bullets are caught. Suppose that the bullets carry plenty of momentum, thanks to a large value of V, but are so Iight that they represent a negligible transfer of mass from the first cart to the second. Suppose further that they are invisible to an observer standing some distance away. (We might make them extremely small, or perhaps paint them black and set up a dark background.) Then if a brief burst of bullets is fired off from A, we see this cart begin to recoil, ap- parently spontaneously, whilc B remains at rest. Not until a time L/V later does B begin to recoil in the opposite direction. There is, in effect, a breakdown of the equality between action and reaction in such a case. By the time the whole interaction has been completed, with all the bullets reabsorbed in B, we recognize that momentum has ultimately been conserved, but if one restricts attention to the carts it does not appear to be con- served, instant by instant, during the interaction. The above example may seem artificial because, after all, we could save the action-reaction principle by looking more closely and observing the bullets at the instant of leaving A or striking B. Nevertheless, it offers an interesting parallel to what are probably the most important delayed interactions — those of electromagnetism. We know that the interaction between two separated charges takcs place via the electromagnetic field, and the propagation of such a field takes place at a speed which, although extremely large, is still finite— the speed of light. The transfer of momentum from one charge to another (resulting, let us say, from a sudden movement of the first charge) involves a time lag equal to the distance between them divided by c. If we looked at the charged particles alone, we would see a sudden change in the momentum of the first charge without an equal and 317 Action, reaction. and impulse opposite change in the momentum of the other at the same time. l There would thus appear to bc a failure of momentum conserva- tion, instant by instant, unless we associate some momentum with the electromagnetic field that carries the interaction. And that is precisely what electromagnetic theory suggests. The picture becomes even more vivid when we introduce the quantiza- tion of the electromagnetic field and recognize that radiation is carried in the form of photons, or light quanta. By the time we have associated a momentum and an energy with each photon individually, we are remarkably close to our mechanical model. There is even a well-developed theoretical description of the static interaction between electric charges in terms of a continual exchange of so-called "virtual photons." In this case, however, since the forces are constant in time, the equality of action and reaction holds good at every instant, and the existence of a finite time for propagating the interaction ceases to be apparent. EXTENDING THE PRINCIPLE OF MOMENTUM CONSERVATION 2 It is perhaps worth amplifying the remarks of the last section a little. We have seen how, in Newton's view, the law of conserva- tion of momentum is closely tied in with the action-reaction idea. There is, however, an alternative approach to simple collisions which loosens this connection and eases the transition to non- Newtonian mechanics. This approach can be defined in terms of a question : What do we actually observe in a collision experiment? The answer is that our observations are purely kinematic ones — measurements of the velocities of the two objects before and after impact. Suppose that two objects, A and B, with initial velocities n,, and u 2 , respectively, collide with one another and afterward have velocities Vj and v 2 - In any individual collision of this type, it is always possible to find a set of four scalar multiples (a) that permit one to write an equation of the following form: <*iui + a2U2 = «:»vi + a 4 \2 (9-12) 'In circumstances where such elfects are important, the very concept of "the same time" — i.e., simultaneity — comes into question. It is just here that Newtonian mechanics itself ceases to be adequate and the revised formulation of dynamics according to special relativity becomes essential. 2 This section may be omitled without destroying the continuity, but it is recommended if you want to see how the bases of classical mechanics can be approached in more than one way. 318 Collisions and conservation laws This as it stands is a quite uninteresting statement. But experi- ments for all sorts of values of u x and u 2 reveal the remarkable result that in every such collision, for two given objects, we can obtain a vector identity by putting <*! = a 3 = a A (a scalar property of A) and a 2 = on = a B (a corresponding scalar prop- erty of B). In other words, the purely kinematic observations on a collision process permit us to introduce a unique dynamical property of each object. Notice that this simple situation ceases to hold if any of the velocities involved become comparable to that of light. In that case it is still possible to construct a vector balance equation in the form of Eq. (9-12), but only if the parameters a are made explicit functions of speed. In fact, we arrive at the relativistic formula for the variation of mass with speed [Eq. (6-3)]. • Let us return now to the results of experiments on collisions at low velocities. The basic statement of these results is that, in the impact of any two given objects, the velocity change of one always bears a constant, negative ratio to the velocity change of the other: V2 — u 2 = — const(vi — m) (9-13) It is precisely because this ratio of velocity changes is found to come out to the same value, whatever the type, strength, or duration of the interaction between the objects, that we can infer that it provides a measure of some intrinsic property of the objects themselves. We define this property as the inertial mass. One can set up an inertial mass scale for a number of objects 1, 2, 3, ... by finding the velocity changes pairwise in such inter- action processes and defining the inertial mass ratio, let us say of objects 1 and 2, by m m |Av[i «II |Av|2 Similarly for objects 1 and 3, rm _ |Av'|i m\ |Av'|.3 and so on. If m t is the Standard kilogram, we then have an operation to determine the inertial masses of any other objects. 'For further discussion, see the volume Special Relaliciry in this series. 319 Exlending ihe principle of momentum conservation We must however, do more. If we let objects 2 and 3 interact, then the ratio m 3 _ |AT"| a m 2 |Av"| 3 must be consistent with the same ratio obtained from the first two measurements. In fact it is, and this internal experimental consistency then allows us to use the values m 2 , m z , ... as mea- sures of the inertial masses of the respective objects. Having set up a consistent measure of inertial mass in this way, we can then rewrite Eq. (9-13) in the form v i ~ »i El (9-14) V2 — U2 mi which when rearranged gives us mim + wi2U2 - mm + »I2V2 (9-15) Thus an equation identical in appearance with the usual mo- mentum-conservation relation emerges, but notice how this con- trasts with the Newtonian analysis. We have used the collision processes themselves to define mass ratios through Eq. (9-13). Once this has been done, the terms in Eq. (9-15) automatically add up to the same total before and after the collision. What we have done here, in effect, is to give primacy of place to momentum conservation. The question as to whether or not action equals reaction does not arise. And this can be very valuable. For when one is confronted with non-Newtonian interactions (i.e., those for which action and reaction are not equal opposite forces at each instant) one faces the problem of how to incorporate them into physics— whether to abandon the law of momentum conservation in its limited form or to extend the idea of momentum and retain the conservation law. The momentum-conservation principle has proved so extremely powerful that the latter course has been chosen, and conservation of linear momentum is a central feature of relativistic dynamics. This way of analyzing the basic results of collision pro- cesses exposes the very intimate relation that exists between kinematics and dynamics. If we changc our description of space, time, and motion, then we can expect that our dynamics must be changed also. This is, in fact, precisely the situation as we make the transition from the kinematics of Galileo and Newton to the kinematics of Einstein's special theory of relativity. 320 Collisions and conservation laws THE FORCE EXERTED BY A STREAM OF PARTICLES The impact of a stream of particles or fluid on a solid surface provides an instructive application of the laws of collision and the conservation of linear momentum. Suppose that a stream of particles, each having mass m and speed v, strikes a block of mass M [Fig. 9-5(a)] and that the particles all lodge in the block. Let the rate (i.e., number per second) at which particles pass through an imaginary fixed plane at right angles to the stream be denoted by R. We know that momentum must be conserved, although if the block were extremely massive one might, in watching the process, receive the impression that the momentum brought in by the particles was simply destroyed, because the velocity acquired is unnoticeably small. Suppose, for example, that the block is stationary when the first particle hits it. At the end of a short time At the number of particles that has arrived is RAt, each carrying momentum mv. If the stream were cut off at this instant, the block and the particles together would continue to move with some constant velocity u (we are assuming that the block is not restrained in any way as far as its horizontal motion is con- cerned). By conservation of linear momentum and mass we then have RmvAt = (M + AM)u where AM = Rm At Fig. 9-5 (o) Stream of individual particles striking a massive object. (b) Initial phase ofbuilding up to a constant average force produced by the particle stream. 321 The forec cxerted by a stream of particles Letting At approach zero, we arrive at two equations that describe the acceleration and the rate of increase of mass of the block: d -§ - Rm - M (9-17) at where we introduce the single symbol n to denote the mean rate of transport of mass in the beam. Clearly, if M is sufficiently large, the block will appear to remain at rest; its displacement and its increase of velocity can remain negligible for some time. But in any event, if the block is stationary at some instant, the force F exerted on it at that instant must be equal to M du/dt. Hence from Eq. (9-16) we have F =* M~r- Rme - po (9-18) at Thus a strcam of particles striking a stationary surface exerts an average fo'rce on it given by Eq. (9-18). The word "average" should be emphasized. Our calculation treats the stream as truly continuous, in the sense that we let At — * in calculating F. But this is a purely mathematical step that does not correspond to the physical reality. The force is produced by the impact of individual discrete particles, and on a sufficiently short time scale we might be able to detect this. Figure 9-5(b) is an attempt to portray the hypothetical results of such observa- tions on the basis of the following very simple model. Suppose that each individual particle, upon striking the surface, undergoes a constant deceleration that brings it to rest in a time T. During this time it must be subjected to a force,/, given by S- T If the force exerted on the block as a result of the arrival of one particle were plotted as a function of time, it would then be a rectangle of height/and width T, as indicated by the small shaded area in Fig. 9-5(b). The force would suddenly come into existence at a certain instant and, a time T later, it would suddenly fail to zero. Suppose now that we consider what happens as a func- tion of time after the beam of particles is first turned on (e.g., by 'The correctness of Eqs. (9-16) and (9-17) depends, in fact, on this condition. If the block has a speed u, the rate at which the particles strike it is reduced from R to R{\ — u/6), and the values of du/dt and dM/dt are reduced accordingly — see Problem 9-13. 322 Collisions and conservation laws opening a shutter that was previously preventing them from reaching the block). Then as successive particles arrive, each adds its contribution / to the total force, which thus rises in an irregular stepwise fashion. However, at a time T after arrival of the first particle, the latter's contribution to the total force would vanish. Thereafter there would be no net increase in the total force, because on the average the earlier particles drop aut of the picture as fast as the new ones appear. The total force thus levels off at some value F but will exhibit statistical fluctua- tions about that mean value. The force will appear to be almost constant if the effects of the individual particles overlap con- siderably in time, as shown in the figure. This corresponds to making the average time interval between successive particles very short compared to the average time that it takes for an individual particle to be brought to rest. We can use the above microscopic picture to recalculate the total force F. It is equal to the force per particle,/, multiplied by the number of particles that are effective at any one instant. This number is equal to the total number of particles that arrive within one deceleration period, T. [Figure 9-5(b) should make this clear.] Since the rate of arrival is R, this number is just RT. Thus we can put F„ = {RT) f = RT™ i.e., F = Rmu as before. It is noteworthy that the average force the beam exerts against the plate (and hence also the average force the plate exerts against the beam of atoms) is quite independent of the actual magnitude of the deceleration that each atom undergoes. The average force is simply equal to the rate of change of mo- mentum of the beam. This force exerted by a stream of particles has been ex- ploited by W. Paul and G. Wessel to measure the average speed of a beam of silver atoms. ' A beam of atoms, evaporated from an "oven," was directed downward onto the pan of a very delicate balance. The force exerted on the balance pan was thus made up of two parts: a steady force, as given by Eq. (9-18), and a 'W. Paul and G. Wessel, Z. Phys. 124, 691 (1948). 323 The force exerted by a stream of particles Force on balance pan Fig. 9-6 Time dependence of a total force due t o the mo- mentum transfer and the tveight of a stream ofpar- ticles striking an object. O -Time Beamturned on at this instant force increasing linearly with time, due to the increasing mass as given by Eq. (9-17) (see Fig. 9-6): ^total = V» + M' The forces involved were equivalent to the weight of a small number of micrograms. The experimental values for an oven temperature of 1363°K (= 1090°C) were approximately as follows: tiv = 3.4 X 10- 7 N M = 5.6 X 10- ,0 kg/sec leading to a value for the average speed, v, of silver atoms at 1363°K: v «= 600 m/sec REACTION FROM A FLUID JET Just as the impact of a stream of particles on a surface produces a push, so the production of such a stream in the first place must cause a force of reaction on the system that gives the stream its momentum. In a normal jet of liquid or gas, the basic granularity of the stream is too fine to be noticed, and any device that sends out such a jet will experience a steady force of reaction, as given by Eq. (9-18). Thus if we imagine a bench test of a rocket engine, for example, with the engine clamped to a rigid structure, then the burnt fuel is thrown out backward with speed vq, as shown in Fig. 9-7(a), and the forward push, P, exerted on the rocket is given by 324 Collisions and conservation laws Fig. 9-7 Sehemat ic diagrams o/a lest-bed arrange- ment of(a) a roeket englne and (b) a jet engine with air intake at the front. P = l*VO One sees this same principle at work on a smaller scale in garden sprinklers, fire hoses, and so on. A jet engine of an aireraft presents a somewhat more com- plicated application of these dynamical results. In this case the air that enters at the front of the engine, and leaves as part of the exhaust gases at the rear, plays an important role in the over-all process of momentum transfer. The main funetion of the fuel that is carried with the plane is to give the ejected gases a high speed with respect to the plane, and most of the moving mass is supplied by the air. It is very convenient to analyze the dynamics of this system from the standpoint of a reference frame in which the engine is instantaneously at rest. If the plane is traveling forward at a velocity v, the air is seen as entering the engine at the front with an equal and opposite velocity, as shown in Fig. 9-7(b). Then, at the rear, all the ejected material has a back- ward velocity v a in this frame. Thus, if air is being carried through the engine at the rate /x ftir (kg/sec) and fuel is being burnt at the rate /x fucI , the total rate of change of momentum defines a total forward foree on the engine according to the following equation: P = Wuel»0 + H\t{O0 ~ ") (9-19) Example. A jet aireraft is traveling at a speed of 250 m/sec. Each of its engines takes in 100 m 3 of air per second, correspond- ing to a mass of 50 kg of air per second at the plane's flying altitude. The air is used to burn 3 kg of fuel per second, and all the gases coming from the combustion chamber are ejected with a speed of 500 m/sec relative to the aireraft. What is the thrust of each engine? Substituting directly in Eq. (9-19) wc have 325 Reaction from a fluid jet P = 3 X 500 + 50(500 - 250) = 1.4 X 10 4 N Four engincs of this type would thus give a total driving force of about 12,000 lb, a more or less realistic figure. The most spectacular manifestation of these reaction forces, at least in man-made systems, is of course the initial thrust from the rocket engincs in a launching at Cape Kennedy (Fig. 9-8). Fig. 9-8 Launching of a Saturn V rocket. (N.A.S.A. photo- graph.) 326 Collisions and conscrvalion laws The following approximate figurcs are based on published data 1 on the first stage of the Saturn V system: Rate of burning 15tons/sec = 1.4 X 10 4 kg/sec Total thrust 7.6 X 10 6 lb = 3.4 X 10 7 N Total mass at liftoff 3100 tons = 2.8 X 10 6 kg Burntime 2.5 min =150sec One can infer from thesc figures that the speed of the ejected gases is about 2500 m/sec and that the initial acceleration is about 2 m/sec 2 . (Remember that in a vcrtical liftoff one must first overcome the downward gravitational puli on the rocket!) By the end of the first-stagc burn, about 2250 tons of fuel have been consumed, and the total mass is down to only about a quarter of the initial value. The analysis of the accelerative process itself, as mass is continually lost, is the subject of the next section. ROCKET PROPULSION This has bccome such a very large subject in recent years that it is clear that we shall do no morc than touch upon the under- lying dynamical principles. For anything like a substantial dis- cussion you should look elsewhere. 2 The fact remains, however, that the operation of any rocket does depend on the basic laws of conservation of momentum, as applied to a system made up of the rocket and its ejected fuel. For simplicity, let us consider the motion of a rocket out i n a region of space where the effects of gravity are sufficiently small to be ignored in the first approxima- tion. Under this assumption, the only force acting on the body of the rocket is the thrust from the ejected fuel. Suppose that the burnt fuel has a speed v n relative to the rocket. Between time t and time l + At, a mass Am of fuel is burnt and becomes sep- arated from the rocket. The situations before and after ejection are shown in Fig. 9-9, where m is the total mass of the rocket plus its remaining fuel at / + At, and o is its velocity at time t. ■NASA Facts: Space Launch Vehicles. 2 Scc, for cxamplc, M. Barrere et al., Rocket Propuhion, Elsevier, Amsterdam, 1960; S. L. Bragg, Rocket Engines, Geo. Newnes, London, 1962; G. P. Sulton, Rocket Propuhion Elements, Wiley, New York, 1963. 327 Rocket propulsion ■MM|mnBMH|BnnH N/ 'V- ■i'» time i (b) f/g. 9-9 Situalions jusi before and jusi after the ejec- tion of an element Am ofmass by a rocket. The ejection of the fuel is a kind of inelastic collision in reverse, since initially the masses m and Am have the same velocity. By conservation of linear momentum we have (w + Am)u = m(v + Av) + Am(v — vo) Therefore, mo + u Am = mo + mAv + o Am — voAm Therefore, mAv = voAm or Ao = v — (9-20) m This equation is not quite exact. (Why?) But as we let At ap- proach zero, the error approaches zero. As long as Au/t?o is much less than unity, Eq. (9-20) is an excellent approximation. Given the initial total mass w, of the rocket plus fuel, and the final mass w/ of the rocket at burnout, the velocity gained by the rocket can be evaluated. If the initial and final velocities of the rocket are i\ and v f , Eq. (9-20) tells us that V — Vi » »o 2 ~ Am The answer could be obtained numerically, by drawing a graph (Fig. 9-10) of l/m against m, and finding the area (shaded) between the limits m/ and m,. This is a pure number, which when multiplied by v gives the increase of velocity. Analytically, if m is used to denote the mass of the rocket (plus its remaining fuel) at any instant, it is more satisfactory to interpret dm as the 328 Collisions and conservation lavvs Fig. 9-10 Craph of l/m versus m. The shaded area giues a relatlve measure of Ihe gain of velocity resulling from a given mass ehange. change of mass of the rocket in a time di. Defined in this way, dm is actually a negative quantity, which when integrated from the beginning to the end of the burning process gives the value of m; - rrii(< 0). In these terms the change of velocity can be written as the following integral in closed form: "/ f dm . (m.\ — Vi = — 00 / — = Poin i — i J mi m \m f J (9-21) You should satisfy yourself that this is indeed equivalent to the final result of the numerical-graphical method described above. Notice that the time does not enter into the calculation at all, although of course it would do so if we wanted to consider the rate of increase of v or the magnitude of the thrust. Notice also that we would be entitled, at each and every instant, to look at the situation from the frame of reference of the rocket itself. In this frame the velocity of the ejected fuel is always just — v , and Eq. (9-20) follows immediately. It is worth examining some of the implications of Eq. (9-21). The first thing to notice is that the gain of velocity is directly proportional to the speed of the ejected gases. Thus it pays to make v as great as possible. The highest values of v attainable through chemical burning processes are of the order of 5000 m/sec, and in practice, because of incomplete burning and other losses, it is hard to do better than about 50% of the ideal theoretical value for a given fuel (cf. the figure of 2500 m/sec for the LOX-kerosene mixture of Saturn V, first stage). These velocities are, of course, very high in ordinary terms, but they are small compared to the velocities that can be given to charged 329 Rocket propulsion particles by electrical acceleration. Hence the interest in develop- ing ion-gun engines, or even using the highest available speed (that of light) by making an exhaust jet of radiation. The trouble with both of these, however, is the very small rate of ejection of mass, which makes the attainable thrust very small. The other main feature of Eq. (9-21) is the way in which the increase of velocity varies logarithmically with the mass ratio. This places rapidly increasing demands on the amount of fuel needed to confer larger and larger final velocities on a given payload. Suppose, for example, that we wanted to attain a velocity equal to the exhaust velocity v , starting from rest. Then, by Eq. (9-21) we have v/ — V( = vo = foln © Therefore, »-. -2.718... m, But to attain twice this velocity, we need to have -fe) " 2 i.c, Uli 2 - . — = e «7.4 m, Table 9-1 presents the results of such calculations in more con- venient form. The last column represents the extra mass needed, TABLE 9-1 v, - v t mi/m, (w< - m,)/m f vo 2.7 1.7 2o 7.4 6.4 3uo 20.1 19.1 4»o 54.5 53.5 as a multiple of the payload. The practical problems of producing very large mass ratios are prohibitive, but the use of multistage rockets (which also have other advantages) avoids this difficulty (see Problem 9-12, p. 359). 330 Collisions and conservation laws A result that may seem surprising at first sight is that there is nothing, in principle, to prevent us from giving a rocket a forward velocity that is considerably greater than the speed v of the exhaust gases. Thus at a late stage in the motion one would see both the rocket and the ejected material moving for- ward with respect to the frame in which the rocket started out from rest. No violation of dynamical principles would be in- volved, and if one made a detailed accounting of the motion of all the material that was in the rocket initially, one would find that the total momentum of the system had remained at zero (as long as the effects of external forces, including gravity, could be ignored). It should of course be emphasized that our whole analysis would hold good, as it stands, only in the absence of gravity and of resistive forces due to the air. COLLISIONS AND FRAMES OF REFERENCE The seventeenth-century investigations of collision processes, by which momentum conservation was established, were chiefly experimental. But one of the men involved — Christian Huygens — applied the spirit of twentieth-century physics in a brilliant analysis of the particular case of two objects with perfect re- bound. He based his argument on symmetry and on the equiv- alence of frames related by a constant velocity. The analysis is simple but of great intrinsic interest. Figure 9-1 1 1 provides a suitably seventeenth-century back- ground to the situation. But though the picture may appear quaint and archaic, the thinking is sharp and fresh. Huygens imagines two equal elastic masses, colliding with equal and opposite speeds ±u. He assumes that they rebound so that each velocity is exactly reversed; this is a symmetry argument. Next, Huygens imagines a precisely similar collision to take place on a boat that is itself moving with speed v relative to the shore. This collision, viewed by a man on terra firma, appears as a stationary mass being struck by a mass with velocity 2v. After the impact, ■From Ernst Mach, The Science of Mechanics, trans. T. J. McCormack, Open Court Publishing Co., La Salle, 111., 1960. The original figure comes from Huygens' Ireatise on impact (ca. 1700). Mach's book (which is extremely readable and uses a minimum of mathematics) is itself a landmark in the discussion of the fundamental principles of physics. His speculations on the origin of inertia were a significant part of the background to Einstein's think- ing about general relativity and cosmology (see Chapter 12). 331 Collisions and frames of rcfcrcnce — m* m i u n i L Fig. 9-11 Huygens' visualization of a n elaslic collision be- tween two equal masses, as shown 'm his book, De Motu Corporum ex Per- cussione (1703). (Reprinled in Vol. 16 ofC. Huygens, Oeuvres Completes, Marlinus Nijhoff, The Hague, Netherlands, 1940.) O m O the first mass has acquired the velocity 2v, and the second mass is stationary. More generally, if the boat has a speed u, different from v, the velocities exchange as follows: Body 1 Body 2 Before impact: u + v u — d Afler impact: u — o u + v Thus Huygens predicts, on theoretical grounds, the results of all possible one-dimensional experiments on the perfectly elastic collision of two identical masses. The feature common to all of them is that the magnitude of the relative velocity has the same value after the collision as it had before the collision. Huygens went even further. Again using a kind of symmetry argument, he deduced a general property of perfectly elastic collisions between unegual masses. If a moving object A strikes an unequal stationary object B, then after the collision they will be moving with velocities v and w as indicated in Fig. 9-12(a). Imagine that this is viewed from a boat moving to the right with velocity w/2. Then before collision the object B is seen moving to the left, with velocity — w/2, as shown in Fig. 9-12(b). Huygens argues that the exact reversal of the motion of B, as seen from this second frame, must also imply the exact reversal of the motion of A in this perfectly elastic process. Thus the final velocity of A must be — (m — w/2) i n this frame. But this velocity is also equal to v — w/2, because the final velocity of A as seen from 332 Collisions and conservation laws Fig.9-12 Elastic collision between two unequal objects as seen (d) from the laboratory frame, and {b) from a frame in which both celocities are simply reversed. (a) (b) Seen from the shore Seen from the boat Before A A O- ""2 After O B B -*o w 2 O^ A or: v - -x B B O*- w 2 the shore is v. Hence we have - (u - w/2) = v - w/2 w — v = u (9-22) Thus Huygens concludes that in the elastic collision of any two objects whatsoever, the magnitude of the relatioe velocity remains unaltered. Notice, however, that something beyond the con- servation of linear momentum is involved here. The extra some- thing is what we have learned to call kinetic energy, and its conservation defines a very special class of collisions. Let us consider this further. KINETIC ENERGY IN COLLISIONS Suppose that a one-dimensional collision takes place as shown in Fig. 9-1 3(a). If this system is effectively free of external in- fluences we have conservation of linear momentum: m\U\. + W2«2 = WlJ-'l + »12 1-" 2 Suppose further that the collision is perfectly elastic, in the sense defined by Eq. (9-22). In the present problem this implies the following relationship: Ml — M2 = t>2 — Di We can solve these two equations for the final velocities, v y and v 2 . The results are »11 — »12 , 2»I2 "i = — — r— "i + r - — «2 1$ = »11 + »»2 2»ii »11 + IW2 «I - »11 + »12 mi — mi »n + »12 U-2 333 Kinetic energy in collisions Using these values of Vi and v 2 , it is a straightforward piece of algebra to arrive at the following result: mivi 2 + W2^2 2 = miui 2 + /W2W2 2 (9-23) You will recognize that this equation, apart from the absence of factors of \ throughout, corresponds to a statement that the total kinetic energy after the collision is equal to the kinetic energy initially. We have arrived at this result on the basis of symmetry considerations and without any explicit mention of forces or work. We shall bring these latter concepts into the picture in Chapter 10, but for the moment we do not need them. Some of the early workers in mcchanics recognized the con- servation property represented by Eq. (9-23) for perfectly elastic collisions, and referred to the quantity mv 2 as the "living force" (Latin, vis vioa) of a moving object. Having, in effect, introduced kinetic energy into the de- scription of collision processes, we shall now develop an impor- tant result that holds good whether or not the total kinetic energy is conserved. Suppose that Fig. 9-13(a) represents an arbitrary one-dimensional collision as observed in a reference frame S. Let us consider this same collision from the standpoint of another frame S' that has the velocity v with respect to S. (In other words, 5 and S' are related by the Galilean transformations.) We then have u\ = «i — V U2 = U2 — v v\ = Pl — B PJ = V2 — o We shall now show that the change of kinetic energy between initial and final states is an invariant — i. e., it has the same value in both frames. In S: ^initial = h m > u l 2 + 4"»2«2 2 #fii.al = i'Wlfl 2 + £'M2U2 2 AK =K,- Ki = (imim 2 + im 2 (.-2 2 ) - (^wiwi 2 + Jm2«2 2 ) In S': ■^'initial = i«l(Ml - V) 2 + \tm(M2 •- V) 2 ^'final = i«l(d - V) 2 + im 2 (V2 - v) 2 AK' = (%mivi 2 + fffttoa 2 ) — (2"»i«i 2 + i"J2«2 2 ) - VftmiVl + W2f2> — (miUi + 7M2«2)] 334 Collisions and conservation laws Fig. 9-13 Arbilrary one-dimensional col- lision as seen in two different f rames related by a velocity v. (a) InS: (b) ms'-. fW o*r 0*r By momentum conservation, for a«;> kind of collision, the com- bination of terms in squarc brackets in AK' is zero. It follows that AK' = AK Thus, if the change of kinetic energy for a given collision process is specified in one frame, it has the same value in all frames. Given a knowledge of this quantity, and the fact of momentum conservation, we can predict the values of u, and v> in any one- dimensional collision process for which the initial velocities w i and u> are specified. The value of AK may be positive, negative, or zero. The first of these corresponds to an explosive process, in which extra kinetic energy is given to the separating particles as a result of the interaction. The second and third possibilities correspond to inelastic and elastic collisions, respectively, in the sense in which we have already been using these terms. THE ZERO-MOMENTUM FRAME The momentum of a particle or of a system of particles is not an invariant; it depends on the frame of reference in which one observes the motion. If, however, one compares the descriptions of the motion in two different inertial frames, related by a con- stant velocity, the difference between the measured values of the momentum of a particle is always a constant vector of magnitude m\, where m is the mass of the particle and v is the velocity of one frame relative to the other. One can always find a reference frame in which the total momentum of any particle vanishes; it is evidently a frame in which the particle is at rest at the instant of time one determines its momentum. One can likewise find a reference frame in which the total momentum of any system of particles is zero, and this zero-momentum frame of reference is of great importance, not only as a convenience for looking at col- lisions and interaetions in general but also, as we shall show 335 The zero-momentum frame Fig. 9-14 Basis of defining the zero- momenlum {center-of- mass) reference frame. shortly, for its dynamical implications. To see how we identify this zero-momentum framc, let us start with a simple example; two particles /n, and m 2 move along the x axis with constant velocities t>i and v% in a frame of reference S (Fig. 9-14). Our task is to find the velocity D of a reference frame S' relative to S such that in S' the total momentum m\v\ + m-iv'i ' s ec l ua ' to zero. Let O' be the origin of S' moving with velocity v relative to O. From Fig. 9-14 we see that the velocities of tn^ and m 2 relative to O are given by v\ = c\ — v v'2 = 02 — u Hence the momentum of the two particles in S' is miv'i + /M2C2 = miL-i — rrtiU + /W202 — mzv and if this is to be zero, we must have (/Ml + m2)v = /Hl»l + OT2l>2 (9-24) Equation (9-24) fixes the velocity of the reference frame S' relative to S but leaves the choice of the position (x) of O' arbi- trary. We use this freedom to choose the location of O' relative to the positions of m, and m 2 as simply as possible. If werewrite Eq. (9-24) in the equivalent form or better as — [(mi + m 2 )x] = — [mixi + m 2 X2] dt di evidently the simplest way of satisfying this is to equate the two 336 Collisions and conscrvalion laws "llg Fig. 9-15 Basis ofdefining the cenler of granit y, not necessarily identical tvilli tJte center ofmass. sets of square brackets. The difference between them must be a m *t constant and we choose the constant equal to zero. We then have for the position of O' in S, _ = misi + m 2 x 2 (9 _ 25) mi + m 2 The origin of a zero-momentum reference frame chosen in this way is called the center of mass of the two-particle system made up of W] and m 2 . You will recognize that Eq. (9-25) also defines what we are accustomed to calling the center of gravity of two objects (cf. Fig. 9-15). In principle, however, these need not be identical points. The center of gravity is literally the point through which a single force, equal to the sum of the gravitational forces on the separate objects, effectively acts. If the values of g at the positions of the two objects were not quite identical, then the forces F gi and F g2 would not be strictly proportional to m! and m 2 , in which case the center of gravity and the center of mass would not quite coincide. The difference is negligible for ordinary purposes, but it should be recognized that the center of mass, as given by Eq. (9-25), is defined without reference to the uniformity or even the existence of gravitational forces. Perhaps you recognized that we have already used the zero-momentum frame in our earlier section on collisions and frames of reference. When Huygens wanted to apply symmetry arguments to col- lisions, this was the frame he started with. But now we have identified it in a much more explicit fashion. The introduction of the zero-momentum or center-of-mass (CM) frame leads to a very important and immensely useful way of analyzing the dynamics of a system of particles. To develop the essential idea in the simplest possible way we shall again consider a one-dimensional system of two particles. In frame S (let us suppose that this is the frame defined by our laboratory) the particles have velocities Vi and v 2 . In the zero-momentum frame (S') they have velocities v[ and v' 2 given by l/l = D] — v V 2 = V 2 — C where 15 is defined by Eq. (9-24). Putting m\ + tn% = M, we can put 337 The zero-momentum frame » - ^±^ = £ (9-26) where P is the total linear momentum, which remains unchanged in any collision proccss: P = ntyvi + HtaD2 = const. (9-27) We now proceed to express thc total kinetic energy K, as mea- sured in the laboratory frame, in terms of the center-of-mass velocity v and the velocities v[ and v 2 of mi and m 2 relative to the CM. We have K = ImM + df + im 2 (o' 2 + vf = (£wiid'i 2 + JTJttPa ) + (miv'i + m 2 v' 2 )o + Kmi + w 2 )0 2 (9-28) Now, by the very definition of the zero-momentum frame, we have miv'i + miv'y = Hence the middle term on the right-hand side of Eq. (9-28) is automatically zero! The first term we recognize as the total kinetic energy K' of the two individual particles relative to the CM; the last term is equal to thc kinetic energy of a single particle of mass M moving with the velocity i; of the CM. Thus we can write K = K' + \Ml n - (9-29) This is such an important result that we shall express it in words also: The kinetic energy of a system of two particles is equal to the kinetic energy of motion relative to the center of mass of the system (the internal kinetic energy), plus the kinetic energy of a single particle of mass equal to the total mass of the system moving with the center of mass. The great importance of this separation of the kinetic energy into two parts — and note well that it works only if the new reference frame is the zero-momentum frame — is that it opens the way to a very powerful and simplifying procedure. We have the possibility of analyzing the internal motion of a system {relative to the CM) without reference to the bodily motion of the system as a whole. (We shall show shortly that the result holds good not only for one-dimensional motions, but in general.) 338 Collisions and conscrvnlion laws One of the implications of Eq. (9-29) is that a certain amount of kinetic energy is locked up, as it were, in the motion of the center of mass. In the absence of external forces the velocity v remains constant throughout the course of a collision process, and the kinetic energy %Md 2 must Iikewise remain unchanged. This means that in the collision of two objects, only a certain fraction of their total kinetic energy, as measured in the labora- tory, is available for conversion to other purposes. The amount of this available kinetic energy is, in fact, identical with K'. We can calculate it with the help of Eq. (9-29): K' = K- $Mo 2 = (imivi + \m2v2) — \{m\ + /»2)0 Substituting for d from Eq. (9-20), this leads by simple algebra to the following result: K' = \ -22?- (c 2 - Bl f (9-30a) 2 mi + rri2 The value of K', as expressed by this equation, is thus the kinetic energy of what can be regarded as being effectively a single mass, of magnitude m l m 2 /(m l + m 2 ), moving at a velocity equal to the relative velocity of the colliding particles. The effective mass is called the reducedmass of the system and is given the symbol y.. Thus we may write Eq. (9-30a) in the more compact form (9-30b) K' = K, where m\m2 u = mi + ni2 Prel = 02 — 1>1 = 02 - -o': For example, if a moving object of mass 2 units strikes a station- ary object of mass 1 unit, two thirds of the initial kinetic energy is locked up in center-of-mass motion, and only one third is available for the purpose of producing deformations, and so on, when the objects collide. COLLISION PROCESSES IN TWO DIMENSIONS In general the velocity vectors of two colliding objects define a plane, and it is therefore important to extend our analysis of collision processes into two-dimensional space. Actually one 339 Collision processes in two dimcnsions Fig. 9-16 (a) Collision seen as occurring in one dimension. (b) As I seen in another refer- I ence frame, the colli- I sion appears two- dimensionat. +M> +.w ■ ■«■■■■■■■M must stand ready to go all the way into three dimensions, because the plane defined by the velocity veetors of two partieles after a collision may not be the same as that defined by the initial velocities. Many collision processes are, however, purely two- dimensional, and for simplicity we shall limit ourselves to such cases in diseussing specific examples, even though the theory applies equally well to three-dimensional problems. One thing that is worth recognizing at the outset is that the analysis of purely one-dimensional problems, of the type we have diseussed so far, can in faet lead to predietions about certain types of two-dimensional collisions, because we can turn a one-dimensional collision into a two-dimensional one by simply changing our point of view. Consider, for example, an imperfectly elastic collision between two identical spheres that approach one another along the y axis with equal and opposite velocities, ±u [see Fig. 9-16(a)]. Suppose that as a result of this collision they recoil back along the y axis with reduced velocities ±o (symmetry requires that these final velocities be equal and opposite). Figure 9-16(b) then shows how the same collision would appear to an observer who had a velocity — w parallel to the x axis. This defines a whole class of oblique collision prob- lems, all of which can be solved with reference to the simple head-on collision of Fig. 9-16(a). One can go even further yet, by viewing the collision of Fig. 9-16(b) from a reference frame that has some velocity parallel to the y direetion. This removes all the apparent symmetry, yct it is basically the same collision. All the collisions for which, at the moment of impaet, the two spheres have a relative velocity ofgiven magnitude along the line joining their centers are dynamically equivalent. 340 Collisions and conservation laws This suggests a vcry important way of simplifying the analysis of collision problems. Imagine yourself in the frame in which the collision is most simply described; this will be the center-of-mass frame, in which by definition the colliding par- ticles approach one another along a straight line with equal and opposite momenta. Solve the problem in this CM frame, and transfer the result to the laboratory frame by means of a Galilean transformation. The grcat simplicity of a collision process as viewed in the zero-momentum frame is illustrated in Fig. 9-17. Since the total momentum in this frame is zero, the particles separate, as well as approach, with equal and opposite momenta, as shown in Fig. 9-17(a). Also the magnitudes of the final momentum vectors are independent of their direction in the CM frame. Thus, as shown in Fig. 9-17(b), the vectors p[/ and p' 2 / can be represented with their tips lying at opposite ends of the diameter of a circle. The relative directions of p{,- and p[/ can be anything, de- pending on the details of the interaction. The relative lengths of these vectors also depend, of course, on the detailed mechanism. In a perfectly elastic collision we have p) = p'i. In a completely inelastic collision we have p' f = 0. (In an cxplosive process, for example rocket propulsion, we have the converse situation, where p'i = 0, p} > 0.) And in atomic, chemical or nuclear reaction processes we may find p} less than, equal to, or greater than p\. To begin with, let us look at one or two examples of perfectly elastic collisions so as to appreciate the beauties of a view from the center-of-mass frame. We are defining an elastic collision Fig. 9-17 (a) An ar- bilrary collision as described by equal and opposite momentum vectors in the CM frame. (b) In the CM frame, the end poinls of the momentum vectors lie on circles. 341 Collision processes in two climcnsions here as one in which the magnitude of the relative velocity is unchanged by the collision. As we have seen, this is equivalent to saying that the total kinetic energy, as well as the momentum, is conserved. The definition in terms of kinetic energy conserva- tion will later become the dominant one. ELASTIC NUCLEAR COLLISIONS The nuclear physicist livcs in two worlds. One is his laboratory, the other is the center-of-mass frame of the particles whose collisions and interactions he is studying. By learning to skip nimbly from one frame to the other he gets the best of both worlds. Let us see how. Proton-proton collisions Figure 9-18 shows a collision between two protons, as recorded in a photographic emulsion. One of the protons belonged to a hydrogen atom in the emulsion and was effectively stationary before the collision took place; the other entered the emulsion with a kinetic energy of about 5 MeV. The most notable feature of the collision is that the paths of the two protons after collision make an angle of 90° with each / / / . Fig. 9-18 Elastic collision between an incident proton and an initially stationary proton in a photo- graphic emulsion. {From C. F. Powell and G.P.S. Occhia- lini, Nuclear Physics in Photographs, Ox- ford Unicersity Press, New York, 1947.) y . ~ «.„.. -%.,..; «...^... \ 342 Collisions and conservation laws othcr. This is true for all such proton-proton collisions, until we get up to energies so high that Newtonian mechanics is no longer adequate to describe the situation. By first looking into the center-of-mass frame we can readily understand this. Let the velocity of the incident proton as observed in the laboratory be v. Then the zero-momentum frame has velocity v/2, and in this frame the protons approach and recede with equal and opposite velocities as shown in Fig. 9-19(a). Suppose one proton emerges from the collision in the direction d', so that the other is at ir — B' on the other side of the line of approach. To get back to the laboratory frame we add the velocity v/2 to each proton, parallel to the original line of motion, as shown in Fig. 9-I9(b). But the triangles ABC and A EF are both isosceles, so the directions 0i and 8 2 of the protons as observed in the laboratory are given by $i = 6' /2 02 = (ir - 6')/2 Therefore, 0i + 6 2 = r/2 Moreover, we can easily find the laboratory velocities of the two protons after the collision, for we have ui = 2(y/2)cos0i = ucosfli v 2 = 2(u/2)cos0 2 = wsinfli Fig. 9-19 (a) Elastic collision between two eaual masses, as seen in the CM frame. (b) Transformation to the laboratory frame, showing a 90° angle between the final velocities. 343 Elastic nuclear collisions We see that in any such collision the total kinetic energy is con- served, because Initial KE = \mv 2 Final KE = £m(t>cosfli) 2 + \m(o cos 2 ) 2 = £/wt; 2 (cos 2 6i + sin 2 0,) = %mv 2 Figure 9-20 shows a similar collision between equal macroscopic objects (billiard balls); it is not perfectly elastic, but it is impres- sively close. Neutron-nucleus collisions In nuclear fission processes, neutrons are ejected with a variety of energies, but the average energy is of the order of 1 MeV. Fig. 9-20 Slroboscopic pholograph of an almost perfectly elastic collision between equal masses. (Front PSSC Physics, D. C. Heath, Lexington, Massachusetts, 1965.) 344 Collisions and conscrvation laws These neutrons are however, most effective in causing further fissions if they are reduced to energies of the order of 10~ 2 or 10 -1 eV (thermal energies). Thus an essential feature of every slow-neutron reactor is a means of slowing down the neutrons. And elastic collisions of neutrons with other nuclei (those com- posing the moderator material of the reactor) do most of the job. Suppose that a neutron of mass m makes an elastic collision with a nucleus of mass M. Let the initial velocity of the neutron in the laboratory be v ; in this frame the struck nucleus will be assumed stationary. Figure 9-21 (a) shows the collision as seen in the center-of-mass frame, which has a velocity v relative to the laboratory frame given by Eq. (9-24): M + m »0 (9-31) If the collision turns the neutron through an angle d' in the zero- momentum frame, its final velocity in the laboratory frame is the vector v shown in Fig. 9-21 (b). The magnitudes of the final velocities, as measured in the laboratory, for any given value of d', are readily calculated. (Exercise: Do this calculation for yourself, and verify that in this case, as with two equal masses, the total kinetic energy, as measured in the laboratory, remains unchanged as a result of the collision.) The biggest energy loss for a neutron as seen i n the lab- oratory frame occurs if it is scattered straight backward (d' = x). In this case we have i;(t) = D - (v - B) = -(to - 2v) Fig. 9-21 {a) Elastic collision between two uneaual masses, as seen in the CM frame. (b) Transformation to the laboratory frame; the angle between the velocities is different from 90°. 345 Elastic nuclear collisions ■«--«*(i-jjF?y M - m For e' = the neutron loses no energy at all. (What sort of a collision is this?) Thus the kinetic energy of the neutron after the collision lies between the following limits: K mtiX = ^mvo Since K max is independent of the mass M of a moderator nucleus, it is the expression for K mKn that tells us what value of M is most likely to lead to the greatest reduction of the average neutron energy. And we see that M = m makes K min equal to zero; we cannot do as well as this for any other value of M, whether it be bigger or smaller than m. Thus if no other considerations were involved, ordinary hydrogen would make the best moderator, since in this case M (the proton mass) is equal to m within about 1 part in 10 3 . Protons, however, also capture slow neutrons rather effectively, thereby making them unavailable for causing further fissions, and it turns out that certain other light nuclei (e.g., deuterium, beryllium, and carbon) offer a better com- promise between moderating and trapping of the fission neutrons. INELASTIC AND EXPLOSIVE PROCESSES We shall turn now to processes in which there may be a net loss or gain of kinetic energy as a result of the collision. In analyzing such processes, it is important to confirm that the change of total kinetic energy is indeed an invariant quantity, and that the result we obtained on the basis of a one-dimensional collision does hold good in general. In order to show this, we first redevelop Eq. (9-29) for two particles moving in arbitrary directions. Let the particles have masses m , and m 2 and velocities vj and v 2 in the laboratory frame (5). Then the velocity, v, of the CM frame (S') is defined by the equation _ = wiivi + >"2V2 (9-33) mi + mz If the velocities of the particles as measured in 5' are v[ and v' 2 , 346 Collisions and conservation laws we then have Vl = v 'i + v v 2 = v' 2 + v We now write down the total kinetic energy in S, using the fact that the kinetic energy $mv 2 of a particle can also be expressed as Jm(v • y), i.e., in terms of the scalar product of the vector v with itself. Thus we have K = £mi(vi • vi) + £m 2 (v 2 • v 2 ) = }«i(Ti + v) • (v| + v) + $m 2 (v' 2 + v) ■ (v 2 + v) Now consider one of these scalar products, using the distributive and commutative laws that apply to the dot products of vectors: (v'i + v) • (v', + v) = v', ■ vi + 2v'i ■ v + v • v = v? + 2v', • v + v 2 Using this result and its counterpart in the second term of the above expression for K, we have K = (Jmii/i 8 + im 2 02) + (miT 7 , + m 2 v 2 ) • v + i(m, + m 2 )D 2 But again we note that, by the definition of the zero-momentum frame, the second term on the right is zero, so that we come back to the simple result of Eq. (9-29): K = K' + £A/B 2 Applying this to the states of a two-particle system before and after collision, we have Ki = K'i + iMV 2 K, = K} + %Md 2 We assume that the total mass remains unchanged and, in the absence of external forces, so does D. Thus we again arrive at the result K/ - Ki = K} - Ki- Q where Q is the amount by which the final kinetic energy exceeds (algebraically) the initial kinetic energy. The actual value of Q may, of course, be negative. The results of the above analysis can be extended, with virtually no modification, to such processes as nuclear reactions, 347 Inelastic and explosive processes in which the actual identity of the particies in the final state may be quite different from what one has at the beginning. Suppose, for example, that there is a collision between two nuclei, of masses m 1 and m 2 , which react to produce two different nuclei, of masses m 3 and m 4 . Then we can write the following statements of conservation : Mass: my + /W2 ■» ms + »14 = M Momentum: /wivi + W2V2 = W3V3 + mi\4 = Mv wiv'i + »i2V2 = W3V3 + mnv'n = Thus the initial and final kinetic energies can be written as follows: Ki = im^f + \m 2 {o' 2 f + £MD 2 K f = ImM? + kmiMf + \Mo 2 with K f - Ki = Q. Example: The D — D reaction. One of the most famous nuclear reactions (and an important one for the process of energy generation by nuclear fusion) is the reaction of two nuclei of deuterium (hydrogen 2) to form a helium 3 nucleus and a neutron ' : ?H + ?H -» 2He + i/i + 3.27 MeV The 3.27 MeV represents the extra amount of kinetic energy, Q, made available because the masses of the product particies (their rest masses, to be precise) add up to a little less than the masses of the initial particies; the total energy, including mass equiv- alents, remains constant, of course. Suppose now that a deuteron with a kinetic energy of 1 MeV strikes a stationary deuteron. What is the final state of affairs as viewed in the CM frame? (See Fig. 9-22.) First, let us cal- culate the velocity D. The mass of a deuteron is about 2 amu, or about3.34 X 10" 27 kg. Now, 1 MeV = 1.6 X KT 13 J. There- forc, D , - (HiJ e 1.0 X 10 7 m/sec 'In this equation, the subscript before the letter for a given nucleus denotes the number of protons and the superscript shows the total number of nucleons. D and ?H both stand for a deuteron. 348 Collisions and conservation laws Before m, O m, O Fig. 9-22 (o) Reac- tion process, in which the collision ofthe particles m i and m 2 leads to the formation oftwo different par- ticles, m z and m,. (6) Same process as seen in the CMframe. After m, After O'"» In Laboratory frame In CM frame (a) (b) Since m 2 = ffli and m 2 is initially at rest, we have D = d i/2 = 0.5 X 10 7 m/sec In the CM frame the deuterons have equal and opposite velocities of magnitude equal to C. Hence we have Kinitial = 2 X %miC 2 = £/HU>l 2 i.e., Kfnitiai = hKx = 0.5 MeV [We see here a particular application of Eq. (9-30). If a moving object collides with a stationary one of equal mass, only half the initial kinetic energy is available for their relative motion in the CM frame.] Now consider the result of the nuclear reaction. The final total kinetic energy i n the CM frame is given by Kfinal = Kinitiol + 3.27 McV = 3.77 MeV = 6.03 X 10" ,3 J This is partitioned between the 3 He and the neutron in such a way that the momenta are numerically equal. Denoting the masses and velocities ofthe 3 He and the neutron as m 3 and v' 3 and m 4 and V4, respectively, we have »»3v'j + W4V4 = «a ,./ «4 349 Inelastic and cxplosivc processes Then = |m 3 (,'3) 2 + ^(^)W 1 W3( ffl3 + W I4) , , ,.2 = 2 ^ «> Putting m 3 « 3 amu and m 4 « 1 amu, this gives us and so U4 (= 3»!) « 2.3 X 10 7 m/sec. We thus have a full picture of the final situation as viewed in the CM frame for any specified direction 9± of the outgoing neutron. To go back to the laboratory frame we have simply to add the CM velocity v to each of the vectors V3 and v 4 . The great advantage of using the CM frame in this way is that, regardless of the final directions as specified by 4 , the magnitudes of V3 and v^ always have the same values, whereas in the laboratory frame v 3 (and also v 4 ) has a different magnitude for each direction. This does not mean, however, that it is always desirable or necessary to go into the CM frame. For example, one may wish to answer the question: What is the speed of a neutron emitted at some given direction 4 in the laboratory with respect to the initial direction of a deuteron beam? In such a case it is easiest to work directly from the equations for energies and momenta as measured in the laboratory: ft + Q = * 8 + K< (934) Pl = P3 + P4 (In the first equation Q represents the amount by which AT flnal differs from Alitlah so that in the example we have just con- sidered we have Q = +3.27 MeV. Q may be positive, negative, or zero; the last of these represents an elastic collision.) Note that Eq. (9-34) represents three independent equations (one for kinetic energy, and two for momentum — treating this as a two-dimensional problem). I n the final state there are four unknowns: a magnitude and a direction for each of the vectors v 3 and v 4 . The situation is indeterminate unless we put in one more piece of information, as for example the direction of one of the particles. Q is here taken to be a known quantity (as are K x 350 Collisions and conservation laws and p O, but one could deduce it from a complete set of measure- ments of v 3 and V4. WHAT IS A COLLISION? We are so used to associating the word "collision" with some abrupt, violent event that it may be well to point out that the results we have developed can be applied to other, quite gentle interactions. The only essential features are these: 1. That the interaction is confined, for all practical purposes, within some limited interval of time, so that it can be said to have a beginning and an end. 2. That over the duration of the collision, the effect of any external forces can be ignored, so that the system behaves as though it were isolated. Figure 9-23 shows stroboscopic photographs of a collision between two frictionless pucks carrying permanent magnets, mounted vertically so as to repel each other. There is no contact in the usual sense, but this is certainly a collision within the Fig. 9-23 (a) A "soft" collision, with no contact in the ordinary sense, occurring between two objects (2:1 mass ratio) carrying permanent magnets. (b) The identical collision photographed by a camera moving with the center ofmass of the two objects. (From the PSSC film, "Moving with the Center of Mass", by Herman R. Branson, Education Development Center Film Sudio, Newton, Mass., 1965.) (a) (b) 351 Whal is a collision? physicist's meaning of the term. One can see that the velocities are effectively constant in magnitude and direction except over the limited region of close approach of the objects. INTERACTING PARTICLES SUBJECT TO EXTERNAL FORCES Having discussed the conservation of linear momentum, and the general relation of force to rate of change of momentum, we can now consider the motion of a system of interacting particles that are not free of external influence. This is, of course, a very im- portant extension of our ideas, because in practice a system is never completely isolated from its surroundings. It will suffice to look at a two-particle system. The ex- tension to any number of particles is quite simple but will be deferred to Chapter 14. Let m, and m 2 be the masses of the two particles, Fi and F 2 the external forces acting on them, and f 2 i and f 12 the internal interaction forces — f 2 i the force exerted on particle 1 by particle 2, and f i2 the force exerted on particle 2 by particle I. Newton's law of motion applied to the particles individually states that Fi + f21 = /wi -7- m (9-35) t/v 2 F 2 + fi2 = nt2 —7- If the interaetions are Newtonian, and we shall consider this the case, then the third law of Newton requires that fl2 + f21 = Adding the two equations (vectorially), we then get Fi + F 2 = mi — + m 2 -j- or Fi + F 2 = -- (mi vi + /n 2 v 2 ) (9-36) at in which the internal forces f 1 2 and f 2 1 have disappeared. This equation states that the resultant of all the external forces acting on the system equals the rate of change of the total veetor mo- mentum of the system. 352 Collisions and conservation laws We can express this result in another, very compact way by introducing the concept of the center of mass of the system. If the individual positions of the two particles are given by the vectors r! and ra, drawn from some origin, the position of the center of mass is given by . _ /rnri + w 2 r 2 (g _ 37) m\ + rri2 This is the three-dimensional analogue of (9-25) and corresponds to the vector velocity of the center of mass as already defined by Eq. (9-33), so that »»lVl + /W2V2 = Mv where M (= m\ + m 2 ) is the total mass of the system. Ac- cordingly, we have F, + F 2 = F = M ^ (9-38) di and this proves the result we are after: The motion of the center of mass of a system of two particles is the same as the motion of a single particle of mass equal to the total mass of the system acted on by the resultant of all the ex ternal forces which act on the individual particles. The implications of this result are significant. First, it suggests that a fundamental method for treating the motion of a system of particles is to analyze its motion as the combination of (1) the motion of its center of mass and (2) the motions of the particles relative to their center of mass. The latter motion, the internal motion of the system, is one of zero momentum, as we saw earlier. Furthermore, Eq. (9-38) allows us to treat some aspects of the motions of extended objects by the laws of dy- namics for a simple particle. In particular, when an extended object moves in translation, i.e., when there is no motion of any particle in the object relative to the center of mass, Eq. (9-38) tells the whole story. We shall return to such questions in Chap- ter 14. Incidentally, Eqs. (9-36) and (9-38) also provide a basis for a criterion as to whether or not a system of colliding particles is effectively isolated. The conservation of momentum in a col- lision process holds good only to the extent that the effect of any external forces can be ignored. If external forces are indeed present, the duration A/ of the collision must bc so short that 353 Intcracting particles subject to cxternal forces the product F At is negligible. A different way of stating this same condition is that the forces of interaction between the colliding particles must be much greater than any external forces which may be acting. THE PRESSURE OF A GAS As our last example of collision processes and momentum transfers we shall take the calculation of the pressure of a gas, on the basis that this is due to the perfectly elastic collisions of the molecules of the gas with the atoms that comprise the wali of the container. We shall assume that the gas is made up of n particles, each of mass wi . P er urut volume. The most naive possible calculation makes the following assumptions: 1. Ali the particles have the same speed, v. 2. The particles can be regarded as though one third of them were, at any instant, traveling parallel to a given direction in space, with the other two thirds traveling parallel to two other directions, perpendicular to each other and to the first direction. 3. The gas is in a rectangular container with perfectly flat, hard walls. None of these assumptions is in the least realistic as it stands. The first two are certainly false, the choice of a rectangular container is very special, and on an atomic scale the walls are knobbly and sticky, not flat and hard. Nevertheless, the calcula- tion we shall make, using these assumptions, comes very close to yielding the correct result. The reason is that, on the average, with the huge number of particles present in a sample of gas, there is an essential symmetry in the aggregate motion and an effectively exact conservation of the total kinetic energy. An individual molecule may strike the wali, sit there a short while, and jump off again at a different angle with a different speed— perhaps faster, perhaps slower. But on the average, we may treat the collisions as perfectly elastic because the kinetic energy of the gas as a whole neither increases nor decreases with time. On this basis, we make the following very simple calculation. Consider an element of area AA in one wali of the containing vessel (Fig. 9-24). Resolve the motions of the molecules so that the normal to AA is one of the three mutually perpendicular directions along which the molecules are assumed to move. Any 354 Collisions and conservalion laws [«•— t>Af C 3 Fig. 9-24 Simplest possible approach to calculating the pressure of a gas. D uring a time At the elemen! of AA wali AA can be reached only by molecules that lle inilially within the cylinder oflengtk v At. one molecule approaches the wali with momentum m v and re- coils with momentum — m Q v; it thus provides an impulse 2moU. How many molecules strike A A in a time Ali If we assume that all of them have the same speed v, molecules farther than d At from the wali cannot arrive until after the time Ar has elapsed. Thus our attention is limited to molecules within a cylinder of cross section AA and length v A/. There are n molecules per unit volume. Hence number of molecules within cylinder = nv Ar AA But of these, only one sixth are moving, not just along a line perpendicular to AA, but specifically approaching AA instead of receding from it. This, then, gives us number of impacts with AA in A/ = \no At AA total impulse communicated to AA in Ar = 2mov X 5/J0 A; AA = $nmov 2 AtAA But the average force AF exerted on AA is the impulse divided by Ar. Hence AF = ^nmov 2 AA The mean force per unit area, exerted normal to the containing wali, is what we call the pressure, p. ' Thus we arrive at the result P = ^ = \ nr »o» 2 (9-39) Since nm is the total mass of gas per unit volume, which is its density p, we can alternatively put P = hpv 2 or "© 1/2 (<M0) 'In this section, p always refers to pressure, not momentum. 355 The pressure of a gas Thus, if our calculation is justified, we can infer the speed of the invisible molecules from perfectly straightforward measure- ments of the bulk properties of the gas. Takung nitrogen at ordinary atmospheric pressure and room temperature, for ex- ample, we have p w 10 5 N/m 2 = 10 kg/m-sec 2 p « 1.15 kg/m 3 Therefore, v « 500 m/sec How defensible is Eq. (9-39)? A more careful calculation is clearly called for. We have seen experimental evidence (Chap- ter 3) that the molecules of a gas at a given temperature have a wide spread of speeds. So the v 2 in our formula should be re- placed by some kind of an average squared speed, u 2 ,. And certainly we could not, on the strength of our own calculation, place much faith in the numerical factor 3 — although the rigorous theory confirms it, in fact. An acceptable treatment of the prob- lem must investigate the consequences of having molecules approaching the wali in quite arbitrary directions; it is only an accident that this does not change the result in detail. (It would, if we considered the numbers of particles striking AA, instead of the force they exert.) Nevertheless, the simple analysis that we have presented gives us a remarkably useful beginning for the understanding of bulk properties in microscopic terms. And, since the detailed kinetic theory of gases is not our present con- cern, we shall rest content with that. THE NEUTRINO At the beginning of this chapter we pointed out that any con- servation law or conservation principle in physics is provisional, but that if, in the face of apparent failure, it is finally vindicated, then its status may be greatly strengthened. The most dramatic success of the conservation laws of dynamics took place in connection with the neutrino — that elusive, neutral particle emitted in the process of radioactive beta decay. The prediction of its existence stemmed from an apparent nonconservation of energy and angular momentum, but perhaps the most beautiful and direct dynamical evidence for it is furnished by the apparent nonconservation of linear momentum. 356 Collisions and conservation laws PROBLEMS Fig. 9-25 Evidence for the neutrino. The visible tracks of the electron and the recoiling lithium 6 nucleus in the beta decay of helium 6 in a eloud- chamber are not collinear. [Front J. Csikai and A. Szalay, Soviet Physics JETP, 8, 749 (1959).] The situation can be simply stated as follows: It is known that the process of beta decay involves the ejection of an electron from a nucleus, as a result of which the nuclear charge goes up by one unit (if the electron is an ordinary negative electron). If no other particles were involved, the process could be written A-»B + e~ where A is the initial nucleus and B the final nucleus. If A were effectively isolated, and initially stationary, our belief in linear momentum conservation would lead us to predict that, whatever the direction (or energy) of the ejected electron, the nucleus B would inevitably recoil in the opposite direction. Any departure from this, regardless of all other details, would demand the involvement of another particle. Figure 9-25 shows a cloud-chamber photograph of the beta decay of helium 6. The decay takes place at the position of the sharp knee near the top of the picture. The short stubby track pointing in a "northwesterly" direction is the recoiling nucleus of lithium 6; the other track is the electron. There must be another particle — the neutrino — if the final momentum vectors are to add up to have the same resultant — i.e., zero — as the initially stationary °He nucleus. It fails to reveal itself because its lack of charge, or of almost any other interaction, allows it to escape unnoticed — so readily, in fact, that the chance would be only about 1 in 10 12 of its interacting with any matter in passing right through the earth. 9-1 A particle of mass m, traveling with vclocity vq, makes a com- pletely inelastic collision with an initially stationary particle of mass 357 Problcms M. Make a graph of the final velocity o as a function of the ratio m/ M from m/ M = to m/ M = 10. 9-2 Consider how conscrvation of linear momentum applies to a ball bouncing off a wali. 9-3 A mouse is put into a small closed box that is placed upon a table. Could a clever mouse control the movements of the box over the table? Just what maneuvers could the mouse make the box per- form? If you were such a mouse, and your object were to elude pursuers, would you prefer that the table have a large, small, or negligible coefficient of friction? 9-4 In the Phncipia, Newton mentions that in one set of collision experiments he found that the relative velocity of separation of two objects of a certain kind of material was five ninths of their relative velocity of approach. Suppose that an initially stationary object, of mass 3wo, of this material was struck by a similar object of mass 2/wo, traveling with an initial velocity Do- Find the final velocities of both objects. 9-5 A particle of mass mo, traveling at speed do. strikes a stationary particle of mass 2mo. As a result, the particle of mass mo is deflected through 45° and has a final speed of uo/2. Find the speed and direc- tion of the particle of mass 2/wo after this collision. Was kinetic energy conserved ? 9-6 Two skaters (A and B), both of mass 70 kg, are approaching one another, each with a speed of 1 m/sec. A carries a bowling ball with a mass of 10 kg. Both skaters can toss the ball at 5 m/sec relative to themselves. To avoid collision they start tossing the ball back and forth when they are 10 m apart. Is one toss enough? How about two tosses, i. e., A gets the ball back? If the ball weighs half as much but they can throw twice as fast, how many tosses do they need? Plot the entire incident on a time versus displacement graph, in which the positions of the skaters are marked along the abseissa, and the advance of time is represented by the increasing value of the ordinate. (Mark the initial positions of the skaters at x = ±5 m, and include the space-time record of the ball's motion in the diagram.) This situation serves as a simple model of the present view of interaetions (repulsive, in the above examplc) between elementary particles. An attractive interaetion can be simulated by supposing that the skaters exchange a boomerang instead of a ball. [Thcse theorctical models were pre- sented by F. Reines and J. P. F. Sellschop in an artiele entitled "Neutrinos from the Atmosphere and Beyond," Sci. Am., 214, 40 (Fcb. 1966).] 9-7 Find the average recoil foree on a machine gun firing 240 rounds (shots) per minute, if the mass of each bullet is 10 g and the muzzle velocity is 900 m/sec. 358 Collisions and conservation laws 9-8 Water emerges in a vertical jet from a nozzle mounted on one end of a long horizontal metal tube, clamped at its other end and thin enough to be rather flexible. The jet rises to a height of 2.5 m above the nozzle, and the rate of water flow is 2 liter/min. It has been previously found by static experiments that the nozzle is de- pressed vertically by an amount proportional to the applied force, and that a mass of 10 g, hung upon it, causes a depression of 1 cm. How far is the nozzle depressed by the reaction force from the water jet ? [This problem is based on a demonstration experiment described by E. F. Schrader, Am. J. Phys., 33, 784 (1965).] 9-9 A "standard fire stream" employed by a city fire department delivers 250 gallons of water per minute and can attain a height of 70 ft on a building whose base is 63 ft from the nozzle. Neglecting air resistance: (a) What is the nozzle velocity of the stream? (b) If directed horizontally against a vertical wali, what force would the stream exert? (Assume that the water sprcads out over the surface of the wali without any rebound, so that the collision is ef- fectively inelastic.) 9-10 A helicopter has a total mass M. Its main rotor blade sweeps out a circle of radius R, and air over this whole circular area is pulled in from above the rotor and driven vertically downward with a speed i>o- The density of air is p. (a) If the helicopter hovers at some fixed height, what must be the value of to? (b) One of the largest helicopters of the type described above weighs about 10 tons and has R ^ 10 m. What is to for hovering in thiscasc? Take p = 1.3 kg/m 3 . 9-11 A rockct of initial mass Mo ejects its burnt fuel at a constant rate \dM/dt\ = y. and at a speed po relative to the rocket. (a) Calculate the initial acceleration of the rocket if it starts vertically upward from its launch pad. (b) If vo = 2000 m/sec, how many kilograms of fuel must be ejected per second to give such a rocket, of mass 1000 tons, an initial upward acceleration equal to 0.5 g? 9-12 This rather complicated problem is designed to illustrate the advantage that can be obtained by the use of multiple-stage instead of single-stage rockets as launching vehicles. Suppose that the pay- load (e.g., a space capsule) has mass m and is mounted on a two-stage rocket (see the figure). The total mass — both rockets fully fueled, plus the payload — is Nm. The mass of the second-stage rocket plus the payload, after first-stagc burnout and separation, is nm. In each stage the ratio of burnout mass (casing) to initial mass (casing plus fuel) is r, and the exhaust speed is vo- (a) Show that the velocity i»j gained from first-stage burn, 359 Problem* c C c K A>m starting from rest (and ignoring gravity), is given by vi = v ln A' r /V + «(1 - r). (b) Obtain a corresponding expression for the additional velocity, V2, gained from thc second-stage burn. (c) Adding u i and i>2, you have the payload velocity v in terms of N, n, and r. Taking A' and r as constants, find the value of n for which f is a maximum. (d) Show that the condition for v to be a maximum corresponds to having equal gains of velocity in the two stages. Find the maximum value of o, and verify that it makes sense for the limiting cases de- scribed by r = and r = 1 . (e) Find an expression for the payload velocity of a single-stage rocket with the same values of A', r, and vo- (f) Suppose that it is desired to obtain a payload velocity of 10 km/sec, using rockets for which vo = 2.5 km/sec and r = 0.1. Show that the job can be done with a two-stage rocket but is im- possible, however large the value of N, with a single-stage rocket. (g) If you are ambitious, try extending the analysis to an arbi- trary number of stages. It is possible to show that once again the greatest payload velocity for a given total initial mass is obtained if the stages are so designcd that the velocity increment contributed by each stage is the same. 9-13 A block of mass m, initially at rest on a frictionless surface, is bombarded by a succession of particles each of mass bm (<<C m) and of initial speed t>o in the positive x direction. The collisions are per- fectly elastic and each particle bounces back in the negative x direction. Show that the speed acquired by the block after the «th particle has struck it is given very nearly by v = i>o(l — e" an ), wherea = 2Sm/m. Consider the validity of this result for an « 1 as well as for an — » oo . 9-14 Newton calculated the resistive force for an object traveling through a fluid by supposing that the particles of the fluid (supposedly initially stationary) rebounded elastically when struck by the object. (a) On this model, the resistive force would vary as some power, n, of the speed o of the object. What is the value of n'! (b) Suppose that a flat-ended object of cross-sectional area A is moving at speed v through a fluid of density p. By picturing the fluid 360 Collisions and conservation laws as composed of n particles, each of mass m, per unit volume (such that nm = p), obtain an explicit expression for the resistive force if each particle that is struck by the object recoils elastically from it. (c) If the object, instead of being flat-ended, were a massive sphere of radius r, traveling at speed v through a medium of density p, what would the magnitude of the resistive force be? The whole cal- culation can be carried out from the standpoint of a frame attached to the sphere, so that the fluid particles approach it with the velocity -v. Assume that in this frame the fluid particles are reflected as by a mirror— angle of reflection equals angle of incidence (see the figure). You must consider the surface of the sphere as divided up into circular zones corresponding to small angular increments dd at the various possible valucs of 0. 9-15 A particle of mass mi and initial velocity «i strikes a stationary particle of mass mi. The collision is perfectly elastic. It is observed that after the collision the particles have equal and opposite velocities. Find (a) The ratio 012/011. (b) The velocity of the center of mass. (c) The total kinetic energy of the two particles in the center of mass frame expressed as a fraction of %m\Ui 2 . (d) The final kinetic energy of 011 in the lab frame. 9-16 A mass /«1 collides with a mass 012. Define relative velocity as the velocity of 011 observed in the rest frame of 0*2. Show the equivalence of the following two statements : (1) Total kinetic energy is conserved. (2) The magnitude of the relative velocity is unchanged. (It is suggested that you solve the problem for a one-dimensional collision, at least in the first instance.) 9-17 A collision appa ratus is made of a set of n graded masses sus- pended so that they are in a horizontal line and not quite in contact with one another (see the figure). The first mass is jmo, the second is f 2 mo, the third / 3 0!o, and so on, so that the last mass hfma. The first mass is struck by a particle of mass 010 traveling at a speed vo. This produccs a succession of collisions along the line of masses. 361 Problem s "». O o6666666ii A m fm„ fm„ (a) Assuming that all the collisions are perfectly elastic, show that the last mass flies off with a speed o„ given by 1 + /, i>o (b) Hence show that, if / is close to unity, so that it can be written as 1 ± e (with e <JC 1), this system can be used to transfer virtually all the kinetic energy of the incident mass to the last one, even for large n. (c) For / = 0.9, n = 20, calculate the mass, speed, and kinetic energy of the last mass in the line in terms of the mass, speed, and kinetic energy of the incident particle. Compare this with the result of a direct collision between the incident mass and the last mass in the line. 9-18 A 2-kg and an 8-kg mass collide elastically, compressing a spring bumper on one of them; the bumper returns to its original length as the masses separate. Assume that the collision takes place along a single line and that you can cause the collision to occur in different ways, each having the same initial energy: Case A: The 8-kg mass has 16 J of kinetic energy and hits the sta- tionary 2-kg mass. Case B: The 2-kg mass has 16 J of kinetic energy and hits the sta- tionary 8-kg mass. (a) Which way of causing the collision to occur will result i n the greater compression of the spring? Arrive at your choice without actually solving for the compression of the spring. (b) Keeping the condition of a total initial kinetic energy of 16 J, how should this energy be divided between the two masses to obtain the greatest possible compression of the spring? 9-19 In a certain road accident (this is based on an actual case) a car of mass 2000 kg, traveling south, collided in the middle of an inter- section with a truck of mass 6000 kg, traveling west. The vehicles locked and skidded off the road along a line pointing almost exactly Southwest. A witness claimed that the truck had entered the inter- section at 50 mph. (a) Do you believe the witness? (b) Whether or not you believe him, what fraction of the total 362 Collisions and conservation laws initial kinetic energy was converted into other forms of energy by thecollision? 9-20 A nucleus A of mass 2m, traveling with a velocity u, collides with a stationary nucleus of mass 10/w. The collision results in a change of the total kinetic energy. After collision the nucleus A is observed to be traveling with speed Pi at 90° to its original direction of motion, and B is traveling with speed Da at angle 8 (sin 6 = 3/5) to the original direction of motion of A. (a) What are the magnitudes of vi and U2? (b) What fraction of the initial kinetic energy is gained or lost as a result of the interaction? 9-21 A particle of mass m and initial velocity u collides elastically with a particle of mass M initially at rest. As a result of the collision the particle of mass m is deflected through 90° and its speed is reduced to «/\/3. The particle of mass M recoils with speed p at an angle 6 to the original direction of m. (Ali speeds and angles are those observed in the laboratory system.) (a) Find M in terms of m, and v in terms of u. Find also the angle 0. (b) At what angles are the particles deflected in the center-of- mass system? 9-22 Make measurements on the stroboscopic photographs of a col- lision of two magnetized pucks (Fig. 9-23) to test the conservation of linear momentum and total kinetic energy between the initial state (first three time units) and the final state (last three time units). 9-23 A particle of mass 2m and of velocity u strikes a second particle of mass 2m initially at rest. As a result of the collision, a particle of mass m is produced which moves off at 45° with respect to the initial direction of the incident particle. The other product of this rearrange- ment collision is a particle of mass 3m. Assuming that this collision involves no significant change of total kinetic energy, calculate the speed and direction of the particle of mass 3/w in the Lab and in the CM frame. 9-24 In a historic piecc of research, James Chadwick in 1932 obtained a value for the mass of the neutron by studying elastic collisions of fast neutrons with nuclei of hydrogen and nitrogen. He found that the maximum recoil velocity of hydrogen nuclei (initially stationary) was 3.3 X 10 7 m/sec, and that the maximum recoil velocity of nitrogen 14 nuclei was 4.7 X 10 u m/sec with an uncertainty of ±10%. What does this tell you about (a) The mass of a neutron ? (b) The initial velocity of the neutrons used? (Take the uncertainty of the nitrogen measurement into account. Take the mass of an H nucleus as 1 amu and the mass of a nitrogen 14 363 Problems nucleus as 14 amu.) 9-25 A cloud-chamber photograph showed an alpha particle of mass 4 amu with an initial velocity of 1.90 X 10 7 m/sec colliding with a nucleus in the gas of the chamber. The collision changed the direction of motion of the alpha particle by 12° and reduced its speed to 1.18 X 10 7 m/sec. The other particle, initially stationary, acquired a velocity of 2.98 X 10 7 m/sec at 18° with respect to the initial forward direc- tion of the alpha particle. What was this second particle ? Was the collision elastic? (In interpreting your results, take account of the fact that these cloud-chamber measurements of speeds and angles are subject to errors of up to a few percent.) 9-26 A nuclear reactor has a moderator of graphite. The carbon nuclei in the atoms of this crystal lattice can be regarded as effectively free to recoil if struck by fast neutrons, although they cannot be knocked out of place by thermal neutrons. A fast neutron, of kinetic energy 1 MeV, collides elastically with a stationary carbon 12 nucleus. (a) What is the initial speed of each particle in the center of mass frame? (b) As measured in the center-of-mass frame, the velocity of the carbon nucleus is turned through 135° by the collision. What are the final speed and direction of the neutron as measured in the lab frame? (c) About how many elastic collisions, involving random changes of direction, must a neutron make with carbon nuclei if its kinetic energy is to be reduced from 1 MeV to 1 keV? Assume that the mean energy loss is midway between maximum and minimum values. 9-27 (a) A moving particle of mass M collides perfectly elastically with a stationary particle of mass m < M. Show that the maximum possible angle through which the incident particle can be deflected is sin -1 (/w/M). (Use of the vector diagrams of the collision in Lab and CM systems will be found helpful.) (b) A particle of mass m collides perfectly elastically with a stationary particle of mass M > m. The incident particle is deflected through 90°. At what angle d with the original direction of m does the more massive particle recoil? 9-28 The text (p. 348) gives an examplc of the analysis of the dynamics of a nuclear reaction between two deuterons. Another possible reac- tion is the following: ?H + fH -♦ !H + ]H + 4.0 MeV In this case the reaction products are a proton and a triton (the latter being the nucleus of the unstable isotope hydrogen 3 or tritium). Suppose that a stationary deuteron is struck by an incident deuteron of kinetic energy 5 MeV. 364 Collisions and conscrvation laws (a) What are the maximum and minimum possible values of the kinetic energy of the proton produced in this reaction? (b) What is the maximum angle (as observed in the laboratory) that the direction of the triton can make with the direction of the incident deuteron? What is its kinetic energy as measured in the lab- oratory when it is emitted in this direction? (This problem can be conveniently handled with the use of amu and MeV as units through- out.) 9-29 A boat of mass M and length L is floating in the water, sta- tionary; a man of mass m is sitting at the bow. The man stands up, walks to the stern of the boat, and sits down again. (a) If the water is assumed to offer no resistance at all to motion of the boat, how far does the boat move as a result of the man's trip from bow to stern? (b) More realistically, assume that the water offers a viscous resistance given by —kv, where A: is a constant and o is the velocity of the boat. Show that in this case one has the remarkable result that the boat should eventually return to its initial position. (c) Consider the paradox presented by the fact that, according to (b), any nonzero value of k, however small, implies that the boat ends up at its starting point, but a strictly zero value implies that it ends up somewhere else. How do you explain this discontinuous jump in the final position when the variation of k can be imagined as con- tinuous, down to zero? For enlightenment see the short and clear analysis by D. Tilley. Am. J. Phys., 35, 546 (1967). 365 Problcms Conservation of Energy as a principle . . . has played a fundamental role, the most fundamental as some would say, in the last half-century of classical physics, and since then in quantum mechanics. . . . Yet it is a curiosity that this, which is thefirst law of thermodynamics, should have so fruitfully watered the stonier, the more austere terrain of mechanics. C. G. GILLISPIE, The Edge of Objedivity (1960) It may be thought strange that we have brought our discussion of the foundations of mechanics sofar with hardly a mention ofenergy. . . . We have indeed had no desire to slight the significance of this concept, but have wished to emphasize that the logical foundation of mechanics is quite possible without it. . . . The guestion at once arises: why then should it have been introduced at all? This is what we wish to discuss. R. B. LINDSAY AND H. MARGENAU, Foundations of Physics (1936) 10 Energy conservation in dynamics; vibrational motions INTRODUCTION of all the physical concepts, that of energy is perhaps the most far-reaching. Everyone, whether a scientist or not, has an aware- ness of energy and what it means. Energy is what we have to pay for in order to get things done. The word itself may remain in the background, but we recognize that each gallon of gasoline, each Btu of heating gas, each kilowatt-hour of electricity, each car battery, each calorie of food value, represents, in one way or another, the wherewithal for doing what we call work. We do not think in terms of paying for force, or acceleration, or momen- tum. Energy is the universal currency that exists in apparently countless denominations; and physical processes represent a conversion from one denomination to another. The above remarks do not really define energy. No matter. It is worth recalling once more the opinion that H. A. Kramers expressed: "The most important and most fruitful concepts are those to which it is impossible to attach a well-defined meaning." ' The clue to the immense value of energy as a concept lies in its transformation. It is consewed— that is the point. Although we may not be able to define energy in general, that does not mean 'See p. 62, where we quoted Kramers' remark in connection with our dis- cussion ofi the concept of time. 367 that it is only a vague, qualitative idea. We have set up quantita- tive measures of various specific kinds of energy: gravitational, electrical, magnetic, elastic, kinetic, and so on. And whenever a situation has arisen in which it seemed that energy had disap- peared, it has always been possible to recognize and define a new form of energy that permits us to save the conservation law. And conservation laws, as we remarked at the beginning of Chapter 9, represent one of the physicist's most powerful tools for organizing his description of nature. In this book we shall be dealing only with the two main cate- gories of energy that are relevant to classical mechanics — the kinetic energy associated with the bodily motion of objects, and the potential energy associated with elastic deformations, grav- itational attractions, electrical interactions, and the like. If energy should be transferred from one or another of these forms into Chemical energy, radiation, or the random molecular and atomic motion we call heat, then from the standpoint of mechanics it is lost. This is a very important feature, because it means that, if we restrict our attention to the purely mechanical aspects, the con- servation of energy is not binding; it must not be blindly assumed. Nevertheless, as we shall see, there are many physical situations for which the conservation of the total mechanical energy holds good, and in such contexts it is of enormous value in the analysis of physical problems. It is an interesting historical sidelight that in pursuing the subject of energy we are temporarily parting company with New- ton, although not with what we may properly call Newtonian mechanics. In the whole of the Principia, with its awe-inspiring elucidation of the dynamics of the universe, the concept of energy is never once used or even referred to! ' For Newton, F = ma was enough. But we shall see how the energy concept, although rooted in F = ma, has its own special contributions to make. We shall begin with the quantitative connection between work and kinetic energy. INTEGRALS OF MOTION In Chapter 6 we briefly presented the basic notion of the work 'As we remarked in Chapter 9, however, some of Newton's contemporaries, particularly Huygens and Leibnitz, did recognize the importance of an energy- like quantity— the "vis viva" mo 2 —\a certain contexts, such as collisions and pendulum motions. 368 Energy conservation in dynamics done by a force acting on an object, producing a corresponding increase in the kinetic energy of the object. We shall now return to this topic and develop it considerably further. Let us again take, for a start, the simple and familiar case of an object of mass m acted on by a constant force F (we shall begin by assuming a straight-line motion for which vector symbols are unnecessary). Then by Newton's law we have do do F = — = m — — ma di dt Let the force act for a time t during which the velocity changes from Vi to v 2 - Then we have Fl = mal = m(o 2 - ui) (10-la) This expresses the fact that the impulse of F is equal to the change of linear momentum. But suppose we multiply the force by the distance over which it acts, instead of by the time. In this case we obtain VI + V2 . Fx = max = ma — / = £m(ar)(Di + u 2 ) = im(v 2 - ui)(ui + Wl) Therefore, Fx = \mv2 2 - J/wi 2 00-2a) This expresses the fact that the work done by F is equal to the change of kinetic energy. We have no reason to declare a prefer- ence between Eqs. (10-la) and (10-2a) as statements of the effect of the force. In fact, they will in general tell us different things. But before amplifying that last remark, let us note that the restriction to constant force and constant acceleration is unneces- sary. For we have F- m d ° Multiplying by dt and integrating gives us /; '2 r »2 F di = m do = m(o 2 - ci) (10-lb) J Vl Multiplying by dx and integrating gives us 369 Integrals of motion £ '*—£%* But dv dx —rdx — —rdo = vav di dt Thus *! Fdx m •a vdv = £j»((?2 — Ci ) (10-2b) The left-hand side of Eq. (10-1 b) is the total impulse of the force, and the left-hand side of Eq. (10-2b) is of course defined as the total work done by the force. Each of these integrals can be repre- sented as the area under the appropriate graph of F against / or x between ccrtain limits (see Fig. 10-1). The general similarity of Eqs. (10-lb) and (10-2b) is, however, deceptive, because force, displacement, and velocity are vector quantities. The result of applying a force of a given magnitude depends very much on its direction relative to the direction of motion of the object on which it acts. Thus in circular motion, as we have seen, a net force is continually applied to an object, the momentum (also a vector) changes continuously, but the mag- nitude of the velocity docs not change at all. The situation becomes clear if we return to the proper vector statement of Newton's law: F = m — dt It is still possible to integrate directly with respect to /: Fig. 10-1 (a) Force that varies linearly with time. The area under the curve mea- sures the total impulse. (b) The same force plotted as afunction of displacement. The area under the curve measures the total work. 370 Energy conservation in dynamics ¥dt = m(v2 — vi) The left-hand side of the expression is an instruction that, if the direction of the force changes with time, we must form the appro- priate vector sum of all the small impulses F A/ applied in suc- cessive time intervals At between t\ and / 2 . That is exactly how Newton himself conceived the action of the varying force to which a planet, for example, finds itself exposed as it moves along its orbit around the sun (cf. Chapter 13). But what about integrating F over the elements of distance along a path, where these elements (Ar) are themselves vectors? This is essentially a physical question, not a mathematical one. We are asking: What is the effect of applying a force at some arbitrary direction to the motion of an object? Figure 10-2 helps to supply the answer. If a force F! is applied at right angles to v for some very short time At, it changes the direction of the velocity without appreciably changing its magnitude. If a force F 2 is applied along the direction of v, it changes the magnitude of v without changing the direction. Thus if we want to fix our attention on changes in the magnitude of v, we should restrict ourselves to forces or force components along the direction of v. If the net force F acts at an angle d to v, we have its component along v given by |F| cos 6 = m |Av| cos d At where Av is the vector change in v, equal to ¥ At/m [cf. Fig. 10-2(c)]. The element of distance traveled during At is given by Ar = vA/ Fig. 10-2 (a) Velocity change due to an impulse perpendicular to v. (b) Velocity change due to an impulse parallel to v. (c) Velocity change due to an arbitrary impulse. 371 Integrals of motion by the definition of the vclocity vector. Hence the force com- ponent along v, multiplicd by the displacement, is given by |F| [Ar| cos d = m |v| |Av| cos 6 The above is a scalar equation; we have suppressed all reference to the effect of F in changing the direction of v and are left with information about the magnitude only. To express this more neatly we use the notation of the "dot product" (or scalar product) of two vectors : a • b = |a; |b| cos e = ab cos e In this notation, we have F • Ar = m(y ■ Av) But we can now play a neat (and valuable) trick. Consider the quantity v • v. This is a scalar, and its magnitude is just the square of |v|, i.e., o 2 simply. But since v 2 = v • v we have, by differentiation, A(u 2 ) = Av ■ v + v - Av = 2v • Av (This last step is possible because the scalar product of two vectors is indepcndent of the order in which the factors come — i.e., the commutative law holds.) Hence v • Av = J A(u 2 ) It follows that we have F-Ar = Jmift) 2 ) So now, integrating over any path that the body may have fol- lowed under the aetion of the force, we find the relation W = \ F- Jt = lm(02 2 - d') (10-3) -' r i which deseribes in general terms the relation between the work done and the change of kinetic energy. Figure 10-3 illustrates what is involved in evaluating the work integral of Eq. (10-3), in going from a point A (r0 to a point B (r 2) in a two-dimcnsional displacement. 372 Enerev conservalion in dynamies Fig. 10-3 Calcula- tion of work done along a given path. Equation (10-3) displays the most important property of work and energy: They are scalar quantities. An object moving vertically with a speed v has cxactly the same kinetic energy as if it were traveling horizontally at this same speed, although its vector momentum would be quite different. This scalar property of energy will be exp!oited repeatedly in our future work. WORK, ENERGY, AND POWER This chapter is chiefly conccrned with dcveloping some general dynamical methods based upon the concepts of work and me- chanical energy. The practical use of these methods will, however, involve numerical measures of these quantities in terms of appro- priate units. The purpose of this section is to introduce some of these units for future reference. Our basic unit, already introduced in Chapter 6, will be the unit of work or energy in the MKS system — the joule: 1 J = 1 N-m = 1 kg-m 2 /sec 2 If the CGS system of units is uscd, the unit of energy is the erg: 1 erg = 1 dyn-cm = 10 -7 J Before going any further, wc shall introduce a seeming diversion — the concept of power, defined as the rate of doing work: power dW dt (For a mechanical system, putting dW = F • t/r, we have power = F ■ v.) Power is a concept (and a quantity) of great practical importance, because the time that it takes to perform some given 373 Work, energy, and power amount of work may be a vital consideration. For example, a small electric motor may be just as capable of driving a hoist as a big one (given, perhaps, a few extra gear wheels), but it may be quite unacceptable because the job would take far too long. Power is essentially a practical engineering concept ; we shall not be using it in our development of the principles of dynamics. But one of our accepted measures of work is often expressed in terms of a unit of power — the watt. In terms of mechanical quantities, 1 W = 1 J/sec i.e., 1 W-sec = 1 J The most familiar use of the watt is, of course, electrical, through the relation watts = volts X amperes, but it is important to realize that it is not a specifically electrical quantity. ' A con- venient energy unit for domestic purposes (especially one's electricity bill) is the kilowatt-hour (kWh): lkWh = 3.6 X 10 6 J In chemical and thermal calculations the Standard unit is the Calorie, defined as the amount of energy required to raise 1 kg ofwaterfrom 15 to 16°C: 1 Cal = 4.2 X 10 :i J In atomic and nuclear physics, energy measurements are usually expressed in terms of the electron volt (eV) or its related units keV (10 3 ), MeV (10 G ), and GeV (10 9 ). The electron volt is the amount of energy required to raise one elementary charge through 1 V of electric potential difference: leV = 1.6 X 10- 19 J Finally, as Einstein first suggested and as innumerable observa- 'The other familiar unit of power is of course the horsepower. About this we shall say only that 1 hp = 746 W, so that (as a very rough rule of thumb) one can say that it takes about 1 kw to drive a 1 hp electric motor. Most of the power levels of practical importance can be conveniently described using power units based on the watt— microwatts (10 " 6 W, typical of very weak radio signals), milliwatts (10~ 3 W), kilowatts (10 3 W), and megawatts (10° W, useful as a unit in generating-plant specifications). It is worth noticing, for comparison, that the sun's power output is about 3.8 X 10" W! 374 Energy conservation in dynamics TABLE 10-1: THE ENERGY OF THINGS - - 10" J - - io^J - - I0»>J 10"J - - I0"J I0M --io-"> J --io-»j --10--J Energy equivalenl of sun's mass (fi = mc') Daily cncrgy oulpui of our galaxy Kinelic energy of earth* orbital molion (relalive lo sun) Daily energy oulpui of sun Kinelic energy of moon's orbital motion (relalive lo earth) Solar energy received per day on earth World uscof cncrgy in 1950 <I0 M J/y) Kinelic energy of a eyelone Daily energy oulput of Hoover Dam Solar energy per day on 2 square miles Burning of 7000 tons of coal Energy released in complete fission of I kg of U 05 Energy equivalent (£ ■ m< *) of I g of malter Energy of explosion of I ton of TNT Energy conlenl of average daily diet One kilowatt-hour Kinelic energy of rifle bullet Oneerg(IO-'J) Energy cquivalcnt (K — («<•') of one atomic mass unil Energy produced by fission (slow neutrons) of one nueleus of U or Pu Energy equivalent (£" = »i< ■*) of one cleclron mass Average energy lo produce one ion pair in air ( - 35 e V) Energy lo break a DNA molecule in two (-0.1 eV) Kinetic energy per molecule ai room temperature (0.025 eV) Energy of a photon a I radio broadeast frequencies Kinetic energy of an electron moving al I m/sec tions have confirmed, there is an equivalence between what we customarily call mass and what we customarily call energy. In classical mcchanics thcsc arc trcated as entirely separate concepts, but it is perhaps worth quoting this equivalence here, so that we have our selection of energy measures all in one place: 1 kg of mass is cquivalent to 9 X 10 16 J For the sake of interest, we show in Table 10-1 some repre- sentative physical cxamplcs involving different orders of magni- tude of energy, all expressed in terms of our basic mechanical unit, the joule. GRAVITATIONAL POTENTIAL ENERGY Wc shall begin with a simple and very familiar problem. An object has been thrown vertically upward and is moving under gravity, losing speed as it goes higher. Let us take a y axis, positive up- ward, and suppose that the object passes the horizontal level y = yi with velocity u t and reaches y = y 2 with velocity v 2 (Fig. 10-4). Thcn the purely kinematic deseription of the motion is given by v 2 2 = m 2 + 2a0> 2 - .yi) with a- -g leading to the familiar result 1 .„2 _ ,,,2 V2 = v\ 2 — 2g(yz - yi) Now let us consider this in terms of work and energy. The changc of the kinetie energy (K) between >>i and y 2 is given by K 2 — K\ = Jmi>2 a — \mv\ z Using the preceding kinematic equation this can be written K 2 — K\ = — mgiyz — yi) Now the quantity — mg is just the constant gravitational foree, ■Remember, we have defined g as a positice number equal to ihe magnilude of the local gravitational acceleration in any units that may be chosen. 376 Energy conservalion in dynamics 4 -yt u% Fig. 10-4 Velocity change associated with vertkal motion under gramty. F„, that acts on the object as it moves up from y\ to y 2 (see Fig. 10-4). Thus we see that the change of K is precisely equal to the work donc by the gravitational force, in conformity with the general rcsult expressed by Eq. (10-3): K 2 - K x = F,0» 2 - yi) We are going to rewrite this equation in a different way, such that the quantities referring to the position y = y\ appear on one side, and the quantities referring to the position y = y 2 appear on the other. We shall express the result in what may seem at first to be a clumsy fashion, but the reason will quickly appear. Our new statement of the rcsult is K 2 + (-F„y 2 ) -Xi + (.-F yi) (10-4) What we have done here is to deliberately frame the mathe- matical statement in such a way that the sum of two quantities has the same valuc at two different positions. That is the formal basis of the statement of conscrvation of cnergy. We define the potential energy U(y) at any given value of y through the cquation l/GO = -F,y G0-5) (thus making U = at y = 0). ' Notice particularly that U(y) is the negalive of the work done by the gravitational force. Sub- stituting this definition of the potential energy into Eq. (10-4) we thus gct E = K 2 + U 2 = tf , + Ui (10-6) where E is the total mechanical energy. Putting F e — —mg in Eq. (10-5), we have the well-known result £/(/,) = mgh for an object at height h above the ground (or other horizontal level that is defined as the zero of potential energy). For this case 'We shall use the symbol U for potential energy throughout this book. The symbol V is widely uscd also, but we shall avoid it here because we so often use t's (large and small) to denote velocities. 377 Gravitational potentiaJ energy Fig. 10-5 Energy diagram for vertical motion above a hori- zontal surface. the energy-conservation statement can be simply represented as in Fig. 10-5. We shall have more to say about such graphs shortly. You will undoubtedly be familiar with another way of inter- preting a potential encrgy such as U(h) in the last equation. It represents exactly the amount of work that we would have to do in order to raise an object through a distance h, against the gravi- tational puli, without giving it any kinetic energy. In order to achieve this, we must supply an exlernal force, F ext ; if this is insignificantly greater than mg, the object will move upward with negligibly small acceleration (F nct « 0), thus arriving at some higher level (Fig. 10-6) with almost zero velocity. 1 If the object Fig. 10-6 Use ofan exterrtal force to move an object in the direction of increasing gravita- tionai potential energy. 'To make it even more precise, wc could apply a force just a shade bigger than mg for a brief time at the beginning, to get the object moving, then change to F„ t exactly equal to mg for most of the trip (the object thus con- tinuing to move upward at constant velocity under zero net force), and finally let F„, become a shade less than mg for a brief time just before the end, so that the object finishes up at rest at height /i above its starting point. 378 Energy conservation in clynamics is subsequently released, then the work done on it by the gravi- tational force is given to the object as kinetic energy (corre- sponding to traveling from y = h to y = in Fig. 10-5). Many devices (pile drivers are a particularly clear example) operate in precisely this way. MORE ABOUT ONE-DIMENSIONAL SITUATIONS K 2 - Ki = I F(x)dx (10-7) Equation (10-6) is a compact statement of the conservation of total mechanical energy in any one-dimensional problem for which the force acting on an object, due to its environment, depends only on the object's position. To derive this energy-conservation result more generally, suppose that the environment supplies a force F(x) that varies with position x in any arbitrary way — but has a unique value at any given value of x. The work done by this force as an object moves from xi to x 2 is given by W= \ F(x)dx Equating this work to the change of kinetic energy, we have In order to cast this into the form of a conservation statement, we introduce an arbitrary reference point x and express the work integral as follows: r*i r 'z r 'i / F(x)dx = / F{x)dx- / F(x)dx (10-8) y i, J x J x Substituting this in Eq. (10-7), we have K 2 + \-J F(x) dx - Ki + 1 - / F(x) dx\ We then define the potential energy U(x) at any point x by the following equation: U(x) - U(x ) = - / F(x)dx (10-9) Jx Notice once again the minus sign on the right. The potential energy at a point, relative to the reference point, is always defined 379 More about one-dimensional situations as the negatioe of the work done by the force as the object moves from the reference point to the point considered. The value of U(x ), the potential energy at the reference point itself, can be set equal to zero if we please, because in any actual problem we are concerned only with differences of potential energy between one point and another, and the associated changes of kinetic energy. In obtaining Eq. (10-9) we took the force as the primary quantity and the potential energy as the secondary one. Increas- ingly, however, as one goes deeper into mechanics, potential energy takcs over the primary role, and force becomes the derived quantity— literally so, indeed, because by differentiation of both sides of Eq. (10-9) with respect to x, we obtain F(x) = - ^ (10-10) This inversion of the roles is not just a formal one (although it does prove to be valuable theoretically) for there are many physical situations in which one's only measurements are of energy differences between two very distinct states, and in which one has no direct knowledge of the forces acting. The electronic work function of a metal, for example, and the dissociation energy of a molecule, represent the only directly observable quantities in these processes of removing a particle to infinity from some initial location. How the force varies from point to point may not be known well — perhaps not at all. We shall often be spelling out the kinetic energy K in terms of m and v, so that the equation of energy conservation in one dimension is written as follows: $mv 2 + U(x) = E (10-11) Suppose we choose any particular value of x. Then Eq. (10-11) becomes a quadratic equation for v with two equal and opposite roots: D(x) = ± (^^)" 2 dO-12) This is the expression of a familiar result, which we can discuss in terms of motion under gravity. If an object were observed to pass a certain point, traveling vertically upward with speed v, then (to the approximation that air resistance could be ignored) it would be observed, a little later, to pass the same point, traveling 380 Energy conservation in dynamics downward at the same speed v. The direction of the velocity has been reversed, but there has becn no loss or gain of kinetic energy. In such a case the force is said to be conseruatiue. We can see from Eq. (10-12) that this result will hold as long as U(x) is a unique function of x. It means that a particle, after passing through any given point at any speed, will be found to have the same kinetic energy every time it passes through that point again. Under what conditions does the force have this conservative property? It will certainly not be conservative if F(x) depends on the direction of motion of the object to which it is applied. Consider, for example, the addition of a resistive force to the gravitational force in the vertical motion of an object. As the object goes upward through a certain point, the net force on it (downward) is greater than F„. After it reaches its highest point and begins moving down, the net force on it (again downward) is less than F„. Hence the net negative work done on it as it rises is numerically greater than the net positive work as it desccnds. Thus on balance negative work has been done and the kinetic energy as the object passes back through the designated point is less than initially. The crucial feature is, indeed, that the net work done by F(x) should be zero over any journey beginning and ending at any given value of x\ only if this condition is satisfied can one define a potential-energy function. It might seem that an equivalent condition is that F be a unique function of position. In one-dimensional situations this is correct. In two- and three- dimensional situations, howevcr, as we shall see later (Chapter 1 1), the condition that F be a single-valued function of r is neces- sary but not sufficient. The condition of zero net work over any closed path defines a conservative force in all circumstances and should be remembered as a basic definition. THE ENERGY METHOD FOR ONE-DIMENSIONAL MOTIONS The use of energy diagrams, such as that of Fig. 10-5, provides an excellent way of obtaining a complete, although pcrhaps quali- tative, picture of possible motions in a one-dimensional system. Frequently the information so obtained suffices for obtaining physical insight into situations for which analytic solutions are complicated or even unobtainable. In fact, even when analytic solutions can be obtained in terms of unfamiliar functions, they often are of little help in revealing the essential physical charac- teristics of the motion. The general scheme is as follows: We 381 The energy meihocl for one-dimensional motions i. i ,— , — . -E, :■< c 6 I„ «, X <f Fig. 10-7 Hypothetical energy diagram for a one-dimen- sional system. plot U(x) as a function of x, and on the same plot draw horizontal lines corresponding to different total energies. In Fig. 10-7 is shown such a potential-energy curve and several values of the total energy. The kinetic energy K of a particle is equal to (E — U), i.e., to the vertical distance from one of the lines of constant energy to the curve U(x) at any point x. For a low energy E u U(x) is greater than E for all values of x; this would simply imply a negative value of K and hence an imaginary value of v. Such a situation has no place in classical mechanics — although it must not be discarded so lightly when one comes to atomic and nuclear systems requiring the use of quantum mechanics. For a higher total energy E 2 , the motion can occur in two regions, between x% and x 4 or between x 7 and x&. These represent two quite separate situations, because a particle cannot escape from one region to the other as long as its energy is held at the value E 2 . One way of seeing this is, of course, in terms of the impossible negative value of K between x 4 and x 7 . But there is another way which is valu- able as an example of how one "reads" such an energy diagram. Suppose that our particle, with total energy £ 2 , is at the point x = x 3 at some instant. Its potential energy U(x 3 ) at this point is equal to the total energy, for this is where the curve of U(x) and the line E = E 2 intersect. Thus the particle has zero 382 Energy conservation in dynami.cs kinetic energy and hence is instantaneously at rest. However, there is a force on it: At x = x 3 , dU/dx is negative and hence F(x 3 ) is positive — i.e., in the +x direction. Thus the particle accelerates to the right. The force on it, and hence its acceleration, decreases as the slope of the U(x) curve decreases, falling to zero at the value of x at which U(x) is minimum. At this point the speed of the particle is a maximum, and as it moves further in the +x direction (with d(J/dx > 0) it now experiences a force in the — x direction. The diagram displays all this information before us, and shows the kinetic energy £ 2 — V continuing to decrease as the particle approaches x 4 . Finally, at x 4 itself, the velocity has fallen to zero — but there is still a force acting in the -x direction. What hap- pens? The particle picks up speed again, traveling to the left, until it reaches x 3 with its velocity reduced to zero. This whole cycle of motion will continue to repeat itself indefinitely as long as the total energy does not decrease. We have, in short, a periodic motion, of which we can discern many of the principal features without solving a single equation— just by seeing what the energy diagram has to tell us. The motion between x 7 and x 8 is likewise periodic. We can dispose of the other possibilities more briefly, having indicated the method. For a still higher energy E 3 , two kinds of motion are possible; either a periodic motion between X2 and x 5 , or the unbounded motion of a particle coming in from large values of x, speeding up as it passes x s , then slowing down and reversing its direction of motion at x 6 , moving ofT to the right and duplicating all the changes of speed on the way in. Finally, for the still larger energy E 4 , the only possible motion is unbounded; a particle coming in from large values of x, speeding up, slowing down, speeding up, slowing down again, and reversing its direc- tion of motion at x lt after which it proceeds inexorab!y in the direction of ever-increasing x. For each of these motions, the speed at any point can be obtained graphically by measuring the vertical distance from the appropriate line of constant energy to the corresponding point on the potential-energy curve. Caution: The curve of U(x) in Fig. 10-7 is almost too graphic. It tends to conjure up a picture of a particle sliding down the slopes and up the peak Iike a roller coaster. Do not forget that 383 The energy method for one-dimensional motions it is a one-dimensional motion that is the subject of the analysis. The vertical scale is energy, and has nothing necessarily to do with altitude. After this general introduction, let us consider some specific examples of one-dimensional motions as analyzed by the energy method. SOME EXAMPLES OF THE ENERGY METHOD Bouncing ball Suppose that a ball, moving along a vertical line, bounces re- peatedly on a horizontal floor. Let us first imagine that there is no dissipation (loss) of mechanical energy, so that this energy remains constant at some value E. We shall use>> to denote the position of the center of gravity (CG) of the ball, and take y = to be defined by the first contaet of the ball with the floor. We shall take this configuration to correspond to U = 0. For y > the potential energy of the ball is given by U(y) = mgy {y > 0) Now y = does not, in any real physical situation, represent the Fig. 10-8 (a) Energy diagram for a ball bouncing ver- lically. (b) Idealizalion of(a) 10 represent a situation in wliicli the impaet at y = is completely rigid bu t in wlrich there is some dissipation of energy at each bounce. 384 Energy conservation in dynamics lowest point rcached by the CG of the ball. The floor does not exert any force on the ball until it (the floor) has been compressed slightly. An equivalent remark can be made about the ball. Thus the ball certainly moves into the region y < 0. As it does this, however, it experiences a positive (upward) force that increases extremely rapidly as y becomes more negative, and completely overwhelms the (negative) gravitational force that exists, of course, at all values of y. This large positive force gives rise to a very steep. increase of U(y) with y for y < [see Fig. 10-8(a)]. For any given value of the total energy, therefore, the ball oscillates between positions y t and y-i as shown in the figure. The motion is periodic — that is, there is some well-defined time T between successive passages in the same direction of the ball through any given point. Now in practice y x may be numerically very small compared to y 2 - For instance, if a steel ball bounces on a glass plate, we might easily have y x of the order of 0.01 cm and y 2 of the order of 10 cm. Thus for many purposes we can approximate the plot of U(y) against y for y < by a vertical line, coinciding with the energy axis of Fig. 10-8(a). This represents the physically unreal property of perfect rigidity — an arbitrarily large force is called into play for zero deformation. However, if we can justifiably use this approximation, then we have a simple quantitative description of the situation. The motion is confined to y > and is defined by i/w, 2 + mgy = E (y> 0) (10-13) The maximum height h is, of course, defined by putting v u = 0: h = — (10-14) mg To find the period T of the motion we can calculate the time for the ball to travel from y = to y = h and then double it. That is, we have the following relation : [ a / dy 2 J »=0 Vy (10-15) because the elementary contribution dt to the time of flight is equal to dy divided by the speed v y at any given point. Now from Eq. (10-13) we have 385 Some cxamples of the energy mcthod O U = ± 2(£ - mgy) "' m *r Taking the positive root, to correspond to upward motion, we have, from Eq. (10-15), /o L2(£ - ««Oj r* r Ti/a r=2/ Us-2 — J * «6- g7o m/ mg)- y]»« We can simplify this by noting that £/wg is just the maximum height h [Eq. (10-14)]. Thus we have r- J*' * /Wo (A->-) 1/2 This is an elementary integral (change the variable to w = h - y) yielding the result You will, of course, recognize the correctness of this result from the simple kinematic problem of an object falling with constant acceleration, and could reasonably object that this is another of those cases in which we have used a sledgehammer to kill a fly. But it is the method that you should focus attention on, and perhaps the use of a familiar example will facilitate this. It should not be forgotten that most motions involve varying accel- erations, so that the Standard kinematic formulas for motion with constant acceleration do not apply. But Eq. (10-15), in which v is defined at any point by the energy equation, can be used for any one-dimensional motion and can be integrated numerically if necessary. Before leaving this example, let us use it to illustrate one other instructive feature of the energy diagram. We know that the total mechanical energy of a bouncing ball does not in fact stay constant but decreases quite rapidly. Although there is little loss of energy while the ball is in flight, there is a substantial loss at each bounce. Figure 10-8(b) shows how this behavior can be displayed on the energy diagram. Starting at the point A, the history of the whole motion is obtained by following the arrows. The successive decreases in the maximum height of 386 Energy conservation in dynamics bounce, and the inevitable death of the motion at y = 0, are quite apparent by inspection of the figure. Mass on a spring There are very many physical systems— not just ordinary me- chanical systems, but also atomic systems, and even electrical ones— that can be analyzed by analogy with a mass on a spring. The reason for this lies in two features: 1. A mass typifies the property of inertia, which has its analogues in diverse systems and which acts as a repository of kinetic energy. 2. A spring represents a means of storing potential energy according to a particular law of force that has its counterparts in all kinds of physical interactions. We have already studied this problem — the problem of the harmonic oscillator — in some detail (in Chapter7) as an application of Newton's law, but it is well worth analyzing the problem again from the standpoint of energy conservation — in part as an illus- tration of this method but chiefly because the description in terms of energy opens the way to a far wider range of situations. Not only does it provide a pattern for the handling of more complex oscillatory problems in classical mechanics; it also supplies the foundation for formulating equivalent problems in quantum theory. Our starting point will again be the restoring force of an ideal spring as described by Hooke's law: F = -kx (10-16) where x is the position of the free end of the spring relative to its relaxed position, k the "force constant" of the spring, measurable in N/m, and the negative sign gives the direction of the force, opposite to the displacement of the free end. No real spring obeys this law over more than a limited range. The propertics of a real spring can be exprcssed by a graph such as Fig. 10-9(a), which represents the force F (x) exerted by the spring as a function of its extension x. Within the linear range the potential energy stored in the spring is, according to Eq. (10-9), U(x) 0- - j o F{x)dx= + J kxdx = '^- (10-17) 387 Sonic cxamplcs of the energy melhod Fig. 10-9 (a) Restor- ing force versus dis- placement for a spring. (b) Potential- energy diagram asso- ciated with (a). (c) Graph ofapplied force versus extension in a stalic deformalion ofa spring. where wc havc chosen U = for x = 0, i.e., when the spring is relaxed. Figure 10-9(b) shows this potential energy plotted against x. Since the potential-energy change can be calculated as the work done by a force F CKl just sufficient to overcome the spring force itself, the increase of potential energy in the spring for any given increase of extension can be obtained as the area, between given limits, under a graph of F cxt against x [sec Fig. 10-9(c)]. F^t can be mcasurcd as the force needed to maintain the spring at constant extension for various values of x. Outsidc the linear region (whose boundaries arc indicated by dashed lines) F(x) can be integrated graphically so as to obtain numerically the potential 388 Energy conservalion in dynamics Compression Fig. 10-10 Mechanical hysteresis. energy for an arbitrary displacement. Before we examine motion under this spring force, let us consider briefly what would happen if the spring employcd in our example were made of lead, for example. As we compress such a spring from its original length, it will exert a force that behaves very much in the way shown in Fig. 10-9(a) and by the line OQ in Fig. 10-10. However, if we now remove the extcrnal agency, the spring will not return to its original length but will acquire a permanent deformation represented by point x r in Fig. 10-10. Clearly, such a spring exerts a force that is not conservative. The value of the force depends not only on the compression but also on the past history. At the value x x of the compression in Fig. 10-10, we find two values of the force, one as the spring is shortened and the other as it is released. This type of behavior is called hysteresis and results in dissipation of mechanical energy. Let us now consider in more detail the way in which the use of the energy diagram helps us to analyze the straight-line motion of a harmonic oscillator. In Fig. 10-11 are shown the potential energy U = \kx 2 plotted against x, and two different total ener- gies, Ei and E 2 . For a given energy E\, as we have already discussed, tne vertical distance from the horizontal line E x to the curve U = %kx 2 for any value of a: is equal to the kinetic energy of the particle Fig. 10-11 Energy diagram for a spring that obeys Hooke's law. 389 Somc examplcs of the energy method at x. This is maximum at x = 0; at this point all the energy is kinetic and the particle attains its maximum speed. The kinetic energy and hence the particle speed decreases for positions on either side of the equilibrium position O and is reduced to zero at the points x = ±.A\. For values of x to the right of +A x or to the Ieft of — A i , (Ei — U) becomes negative ; v 2 is negative and there exists no real value of v. This is the region into which the particle never moves (at least in classical mechanics); thus the positions x = ±A i are turning points of the motion, which is clearly oscillatory. The amplitude A i of the motion is determined by the total energy E\. Since the kinetic energy Ki (=£1 — U) is zero at x = ±Au we have $kAi2 = Ei or ,1/2 - - (f J (10-18) For a larger energy, E 2 , the amplitude is larger in the ratio (E2/E1) " 2 , but the qualitative features of the motion are the same. It is interesting to note that the general character of the motion as inferred from the energy diagram would be the same for any potential-energy curve that has a minimum at x = and is symmetrical about the vertical axis through this point. All mo- tions of this sort are periodic but differ one from the other in detail, e.g., the dependence of speed on position and the de- pendence of the period on amplitude. Suppose that the period of the motion is T. Then for any symmetrical potential-energy dia- gram, we can imagine this time divided up into four equal por- tions, any one of which contains the essential information about the motion. For suppose that, at t = 0, the particle is traveling through the point x = in the positive x direction. Let its velocity at this instant be called v m — it is the biggest velocity the particle will have during its motion. At / = T/4 the particle is at its maximum positive displacement (x = +A in Fig. 10-11), and u = 0. It then retraces its steps, rcaching* = after a further time T/4 and passing through this point with v = — v m . In two further intervals T/4 it goes to its extreme negative displacement (x = —A) and at / = T is once again passing through the point 390 Energy conservation in dynarnics Fig. 10-12 Sinus- oidal variation of velocity with timefor a particle subjected to a restoring force proportional to dis- placement. The mo- tion during the first quarter-period suffices to define the rest of the curve. x = O with v = v m . This sequence will repeat itself indefinitely. Furthermore, knowing the symmetry of the problem, we could construct the complete graph of v against t from a detailed graph for the first quarter-period alone (see Fig. 10-12, in which the basic quarter period is drawn with a heavier line than the rest). In the next section we shall go beyond this rather general examination of the mass-spring system, including nonlinear restoring forces, and shall redevelop the rest of the detailed results that apply to the ideal harmonic oscillator. Before doing that, however, we shall consider one more simple example that illustrates the usefulness of the energy method. Dynamics of a catapult The catapult is an ancient and effective device for launching stones or other missiles with quite high velocities, by converting the potential energy of a stretched elastic cord into the kinetic energy of a mass. I n Fig. 10-13(a) we show the essentials of the arrange- Fig. 10-13 (a) Initial and final stages o f the launching of an object by a catapult, (b) In- termediale state, showing the instan- taneous force acting. 391 Somc examples of the energy method ment. An elastic cord, of natural (unstretched) length 2/ , is attached to fixed supports at the points A and D (A C = CD = / ). An object of mass m is placed at the midpoint of the cord and drawn back to point B. When it is released, the object travels along the line BC, and at the point C it begins its free flight with speed o. Let us assume that the cord develops a tension proportional to its increase of length. Then we can use Eq. (10-17) to calculate the stored energy when the mass is at the initial position, B.- Remembering that there arc two segments of cord, each of initial length / and stretched length l x , we have U = 2 X i*(/i - /o) 2 = Wi - /o) 2 In the idealization that the cord has negligible mass and thus does not drain off any of the elastic energy for its own motion, we can equate the final kinetic energy of the projectile to the initial potential energy as given above. Thus we have \mv* = /c(/i - /o) 2 and so ,1/2 °-(S) (/i - /o) This example displays particularly well the advantage of making the calculation with the help of the scalar quantity, energy, instead of the vector quantity, force. For suppose we wanted to calculate the final velocity of the mass by direct application of F = ma. Then, as indicated in Fig. 10-1 3(b), we should have to consider the state of affairs at an arbitrary instant when the mass was at some point P between B and C such that AP = I. The ten- sion in each half of the cord would be k(l - / ), and the instan- taneous acceleration would be obtained by resolving these tension forces along the line BC: m^- = F x = 2k(l - /o) cos d at This equation for do/dt would then have to be integrated between the points B and C. Or, alternatively, we could calculate the total work done by the varying force F x along the path BC. In either case, this is a far harder road to the final result than is the direct application of conservation of mechanical energy. 392 Energy conservation in dynamics THE HARMONIC OSCILLATOR BY THE ENERGY METHOD We shall now return to the analysis of the oscillatory motion of an object attached to a spring that obeys Hooke's law. The basic energy equation for a mass on a spring with a restoring forcc proportional to displaccmcnt is £/»w 2 + \kx 2 = E (10-19) where E is some constant value of the total energy. Since v = dx/dt, this can be rcwritten as hm (4j?\ + £**2 = E (!0-20) Equation (10-19) already gives us v as a function of x, but to have a full description of the motion we must solve Eq. (10-20) so as to obtain x (and hence v) as functions of r. Our way of dealing with Eq. (10-20) will appeal to the knowledge of trigonometric functions and their derivatives that one develops at an early stage of any calculus course. We start out by dividing the equation throughout by E. Then we get ni (dx\ 2 k_ 2E\dt/ + 2E x2 = 1 (10-21) We notice that this is a sum of two terms involving the square of a variable (x) and the square of its derivative (with respect to t). The sum is equal to 1 . Now we can relate this to a very familiar relationship involving trigonometric functions: If 5 = sin •/>, then ds and W + s2 = cos2 <p + sin2 p = 1 (10-22) Equations (10-21) and (10-22) are exactly similar in form! We must be able to match them, term by term: 2E\dt) W/ 2£ The second of these is satisfied by putting 393 The harmonic oscillator by the energy meihod X = (t)""' - (t)"'*** '">-*> What is ^s? We can find it by evaluating dx/dt by differentiation of both sides of Eq. (10-24) with respect to t: dx = /2EV dt \k) dip cos„- But the first equation of (10-23) is satisfied by putting * m (2g\ V ** m (2£\ U * QM dt \m/ d(f> \m) Comparing these two expressions for dx/dt we find the following condition on <p: J = (* Y" - u (10-25) dt \m/ where w is the angular velocity (also called the circular frequency) that we met in our previous solution of the harmonic oscillator problem (Ch. 7, p. 226), starting from F = ma. Integrating the last equation with respect to t we thus get (p = wt + ipo where <po is the initial phase. Substituting this expression for <p back into Eq. (10-24) then gives us x = fejtj sin (co/ + w>) (10-26) If we note that (2E/k) u2 is equal to the amplitude A of the motion [Eq. (10-18)] we arrive finally at the same equation for x(t) that we found in Chapter 7 [Eq. (7-42)]. [If you have some prior knowledge of differential equations, you may regard our method of solution above as being rather cumbersome. You may prefer to proceed at once to the recogni- tion that Eq. (10-20) leads to the relationship (I) f)' and hence to the following solution by direct integration: dx ,2 2.1/2 di'"* 4 ~ X) 394 Energy conscrvation in clynamics ,dt dx (4» - x2)I/2 ut + (po = sin -1 i -J and so x = A sin (ut + <p ) as before.] Equations (10-25) and (10-26) tell us something very remark- able indeed: The period of a harmonic oscillalor, as typified by a mass on a spring, is complelely independen! ofthe energy or ampli- lude ofthe motion — a result that is not true of periodic oscillations under any other force law. The physical consequences of this are tremendously important. We depend heavily on the use of vibrat- ing systems. If the frequency v (defined as the number of complete oscillations per second, i.e., l/7"or u/2v) varied significantly with the amplitude for a given system, the situation would become vastly more complicated. Yet most vibrating systems behave, to some approximation, as harmonic oscillators with properties as described above. Let us see why. SMALL OSCILLATIONS IN GENERAL There are many situations in which an object is in what we call stable equi1ibrium. It is at rest at some point — under no net force — but if displaced in any direction it experiences a force tending to return it to its original position. Such a force, unless it has pathological properties (such as a discontinuous jump in value for some negligible displacement) will have the kind of variation with position shown in Fig. 10-14(a). The normal resting position is marked as x . This force function can be inte- grated to give the potential-cnergy graph of Fig. 10-14(b). One Fig. 10-14 (a) Varia- tion of force with dis- placement on either side of the equilibrium position in a one- dimensional system. (o) Potential-energy curve associated with (a). 395 Small oscillations in general UU) Fig. 10-15 Potential-energy curve ofFig. 10-14(b) referred to cm origin located at the equilibrium position. can then form a mental picture of the object sitting at the bottom, as it were, of the potential-energy hollow, the minimum of which is at x = x . Now wc can fit any curve with a polynomial expansion. Let us do this with the potential-energy funetion — but let us do it with reference to a new origin chosen at the point x , by putting x = xo + s where 5 is the displacement from equilibrium. The potential- energy curve, now appearing as in Fig. 10-15, can be fitted by the following expansion : U(s) = U + as + ic 2 s 2 + %c 3 s 3 + (10-27) (The numerical faetors are inserted for a reason that will appear almost immediately.) The foree as a funetion of s is obtained from the general rela- tion m = _ _ so that we have F(s) = —Cl — C2S — css 2 ■ ■ ■ However, by definition, F(s) = at 5 = 0; this is the equilibrium position. Hence c { = 0, and so our equation for F becomes F(s) = — C2S — C3S 2 (10-28) Now, whatever the relative values of the constants c 2 and c 3 , there will always be a rangc of values of s for which the term in s 2 is much less than the term in 5, for the ratio of the two is equal to Czs/ci, which can be made as small as we please by choosing s small enough. A similar argument applies, even more strongly, to all the higher terms in the expansion. Hence, unless our po- tential-energy funetion has some very special properties (such as having c 2 = 0) we can be sure that for sufficicntly small oscilla- tions it will be just like the potential-energy funetion of a spring that obeys Hookc's law. We can write 396 Energy conservation in dynamics U(s) m %c 2 s 2 (• 0-29) which means that the effective spring constant A; for the motion is equal to the constant c 2 . We shall discuss a specific application of this analysis — molecular vibration — later in this chapter. THE LINEAR OSCILLATOR AS A TW0-B0DY PROBLEM So far, in all our discussions of potential energy, we have analyzed the problems as though we had a single object exposed to given forces. A statement of our calculation on an object near the earth's surface could well be in the form: The polen tial energy ofan object of mass m raised to a height h above the earth's surface is mgh. Such a statement is perfectly legitimate for situations in which the mass of a particle is very small compared to the mass of the object (or objects) with which it interacts. In such a case, the center of mass of the system is effectively determined by the position of the larger mass. A frame of reference anchored to this larger mass is both a zero-momentum frame and a fixed frame of refer- ence. This is the case for the earth and an ordinary object moving near its surface. It is also the case for interactions between any two objects if one of them is rigidly attached to the earth. One must remember, however, that, strictly speaking, one is analyzing a two-body system (the earth and the object which is raised): mgh is the increase of potential energy of the system when the separation between the earth and the object of mass m is increased by an amount h. I n other words, the potential energy is a property of the two objects jointly ; it cannot be associated with one or the other individually. If one has two interacting particles of comparable mass, both will accelerate and gain or lose kinetic energy as a result of the interaction between them. It is to this basic two-body aspect of the potential-energy problem that we now turn. Suppose that we have two particles, of masses m, and m 2 , connected by a spring of negligible mass aligned parallel to the x axis (Fig. 10-16). Let the particles be at positions x t and x 2 , as shown, referred to some origin O. If the spring is effectively massless, the forces on it at its two ends must be equal and oppo- site (otherwise it would have infinite acceleration) and hence, accepting the equality of action and reaction in the contacts between the masses and the spring, the forces exerted on the masses are also equal and oppositc. Thus, denoting the force 397 The linear oscillator as a two-body problem Fig. 10-16 System of two masses con- nected by a spring, showing ihe separale coordinates and forces. f« "h Wom c m 2 F, 2 i 700 /i/l/t ■ O •«'i .i -Vs exerted on mass 2 by the spring as F x 2 , the force F 2 i exerted on mass 1 by the spring is equal to — F 12 . We shall relate the changes in kinetic energy of the masses to the changes in stored potential energy in the spring. The potential energy of the spring First, suppose that m i moves a distance dx\ while m 2 moves dx 2 - The work done by the spring is given by dW = Fi2dx2 + F2 1 dx\ = Fi2(dx2 — dx\) (sinceF2i = — Fia) - Fi 2 d(,X2 - Xl) Clearly the difference x 2 — x x , rather than x x and x 2 separately, defines the elongation of the spring (and hence the energy stored in it). Let us introduce a special coordinate, r, to denote this: r = X2 — x\ Then dW = F 12 dr (work done by spring) (10-30) The change of potential energy of the spring is equal to —dW. Introducing the potential energy function U(r) we have dU = - F12 dr (r)= - J F 12 dr (10-31) The kinetic energy of the masses Our discussion of two-body systems in Chapter 9 suggests clearly that we should introduce the ccntcr of mass of the system and refer the motions of the individual masses to it. This allows us, as we have seen, to consider the dynamics in the CM frame with- 398 Energy conscrvalion in dynamics out reference to the motion of the system as a whole. By Eq. (9-29) we have K = K' + £A/£5 2 where K' is the total kinetic energy of the two masses as measured in the CM frame. Denoting the velocities relative to the CM by v[ and u' 2 as usual, we have K' = imivf + \mv£ (10-32) We have seen (pp. 338-339) that it is very convenient to express K' in terms of the relative velocity, v„ and the reduced mass, y., of the two particles: V r = V2 — f'l m\m2 M = From the definition of the CM (zero-momentum) frame, we have mio'i + /«2^2 = Using this, together with the equation for v T , we find , w»2 , mi m = ; v, 02 = ; U T f 2 = "r m\ + n\2 "'i + "i2 Substituting these values into Eq. (10-32) one arrives once again at the result expressed by Eq. (9-30a): r.JJHL«l-i P»r* 00-33) 2mi + n\2 We shall be considering the changes of kinetic energy of the masses as related to the work done on them by the spring. On the assumption that no external forces are acting, we have B = const., in which case dK = dK' The motions At this point we can assemble the foregoing results and equate the change of kinetic energy to the work done by the spring. We evaluated dW [in Eq. (10-30)] in terms of laboratory coordinates, although, as we saw (and could have predicted), the result depends only onxa — Xi, which is equal to x' 2 — x[ — both being equal 399 The linear oscillator as a two-body problem to the relative coordinate r. Likewise, as we have just seen, dK = dK'. We can, in fact, put dK' = F\zdr (work done by spring) Integrating, K' = / F l2 dr + const. And now, with the help of Eq. (10-31), we can write this as a statement of the total mechanical energy E' in the CM frame: K' + U(r) = E' (10-34) For the specific case of a spring of spring constant k and natural length r , we can put r = r + s U(s) = $ks2 Also dr ds Thus the equation of conservation of energy [Eq. (10-34)] becomes J* (f) +ito8-i? 00-35) where mim-2 M = mi + m 2 This is exactly of the form of the linear oscillator equation; its angular frequcncy w and its period T are given by -- ($)"* r - 2 *(?r (, °- 36) It is to be noted that the reduced mass /x is less than either of the individual masses, so that for a given spring the period is shorter i n free oscillation than if one of the masses is clamped tight. COLLISION PROCESSES INVOLVING ENERGY STORAGE With the help of the analysis developed in the last section, we can 400 Energy conservation in dynamics gain further insight into certain inelastic or explosive collisions of the type discussed in Chapter 9. We shall introduce the problem by imagining a little me- chanical gadget that could be constructed without much trouble. The gadget is a spring equipped with a buffer that slides along a guide and becomes lockcd iri place if the spring is compressed by more than a certain amount [see Fig. 10-17(a)]. Suppose that this device (assumed to be of negligible mass) is attached to an object of mass m 2 , and that an object of mass m t collides with it. For simplicity, let us take m 2 to be initially stationary, and let mi approach it with a speed «i. If Mi is small, the collision is perfectly elastic. The spring will be compressed a little when m^ strikes it, but it will return to its original unstretched condition, and at this instant the col- lision process comes to an end. Throughout the time that the spring is at all compressed, the mass m 2 is subjected to an accelera- tive force to the right and m i is subjected to a decelerative force to the left. A certain positive amount of work is done onm 2 and an equal amount of negative work is done on m\. The total kinetic energy is the same after the collision as beforehand, but it has been reapportioned. If ui is increased, the situation is finally reached in which the maximum compression of the spring just brings it to the locking position. At this instant mi and m 2 are at rest relative to one another, and because the spring is prevented from pushing them apart again, this is the way they remain. In other words, they would continue to move on as a single composite object of mass mi + m 2 ; the collision has suddenly become completely inelastic. Fig. 10-17 (a) Collision inooking one object with an energy-storage device. (b) Same collision in the center- of-mass frame. The collision is elastic if the total ki- netic energy in this frame is less than the work needed to compress the spring by the crilical amount. 401 Collision processes involving energy storage What defines the critical condition at which the collision changes from elastic to inelastic? We can most easily discuss this from the standpoint of the CM frame [Fig. 10-17(b)]. In this frame the masses have a total kinetic energy given (cf. the previous section) by K' = ivu> T 2 with u = - — — v T = — "i (because uz = 0) m\ + n\2 If we denote the initial kinetic energy of m x in the Lab frame as Ku we can put mi + m2 Now at the critical value of K x all the energy K' is used up in compressing the spring to the Iocking position; this requires a well-defined amount of work equal to the energy, U , stored in the spring in this configuration. Thus we can put K' = m 2 K, = f/o or _ mi + m 2 Uq (10 _ 37) n\2 If U is given, the above equation defines a threshold value ofKu at which inelastic collisions become possible and below which the collisions can only be elastic. If we consider still higher values of K u we obtain situations in which the spring is compressed beyond its Iocking point but then returns to that critical length and stops. During this partial reextension it pushes m i and m 2 apart, giving them a final relative velocity and a kinetic energy of relative motion in the CM frame. The collision is still inelastic but only partially so. A well-defined amount (C/ ) of the originally kinetic energy of the colliding masses has been locked up in the spring, and the dynamics of the collision can be analyzed in these terms. Our analysis in Chapter 9 of inelastic and explosive collisions was, in fact, made precisely in this way. The quantity Q, representing the value of K/ - K that we introduced there (p. 347) is in this case simply equal to -t/„- 402 Energy conservation jn dynamics The purely classical dynamical situation described above is closely paralleled by many collision processes in atomic and nuclear physics, in which an incident particle (e.g., an electron or a proton) strikes a stationary target particle that has various sharply defined states of higher internal energy than its normal "ground state." The sharpness and discreteness of these charac- teristic states is understood in terms of quantum theory, but all that we need here is the knowledge of their existence. If a study is made of the distribution of kinetic energies of the electrons or protons after they have collided with target particles of a given species, the results give information about the excited energy levels involved. Some examples of this kind of analysis are shown in Fig. 10-18. The higher the bombarding energy of the incident particle, the larger the number of excited states that can be stim- ulated and detected in this way. Figure 10-18(a) shows data from the scattering of electrons by helium atoms, and Fig. 10— 18(b) shows some results for the scattering of protons by the nucleus boron 10. In the former case the target particle is so massive that the available kinetic energy in the CM (K') is insignificantly different from the electron energy K\. In the latter case, however, the center-of-mass effect is considerable and one must use Eq. (10-37) to infer the excitation energy U from the threshold value of Ki> One important difference between these atomic or nuclear scattering processes and the classical ones is that perfectly elastic collisions may still occur, with a certain probability, even after the threshold energy for inelastic processes has been exceeded. This is an example of the general feature of quantum mechanics that one only has relative proba- bilities of events when several outcomes are possible. To return briefly to the classical situations once again, one can of course imagine mechanisms that would provide for an increase rather than a decrease of kinetic energy as the result of a collision. One could, for example, have a spring already com- pressed that would be released if a trip mechanism were activated by the initial impact. Such a process — an explosive collision — would, like the inelastic collision with energy storage, require a certain minimum threshold energy to make it go. Something very comparable to this happens with chemical explosive systems that require the firing of a detonator. It is possible to do a quantitative experiment in which a simple percussion cap (as used in a child's toy pistol) is mounted on a mass that is free to recoil, and is deto- nated by the impact of an incident object (e.g., on an air track). 403 Collision processes involving energy storage 180 190 Scattered electron energy, e' a) ">B<0.72 MeV) Elastic scattering peak • w B*(2.15MeV) •"B*(1.74 MeV) A ">B(0) •C(0) ~ nJll 250 300 Relative proton momentum, units of Br (b) .350 Fig. 10-18 Experimental resulls on elastic and inelastic collision pro- cesses, showing Ihe production of characteristic excited states ofsharply defined energy: (a) Scattering of electrons by neutral helium atoms. [AfterL. C. Van Alta, Phys. Rev., 38, 876 (1931).] (b) Scattering of protons by nuclei of boron 10. Some carbon was also present as an unavoidable contaminant. [After C. K. Bockelman, C. P. Browne, W. W. Buechner, and A. Sperdmo, Phys. Rcv., 92, 665 (1953).] 404 Energy conservation in dynainics The threshold kinetic energy required under these conditions is significantly greater than if the cap is mounted on an unyielding support. THE DIATOMIC MOLECULE 1 The diatomic molecule, as typified by such molecules as HC1 or Cl 2 , is a system to which the methods developed in this chapter can be very effectively applied. In such molecules, the atomic nuclei play the role of point masses (which of course they are to an extremely good approximation, being so tiny in comparison to an atomic diameter) and the outer clectron structure of the atoms plays the role of a spring system that can store potential energy. We thus have the physical basis for characteristic vibra- tions of the nuclei along the line joining them, and this will be a possible mode of internal motion of the molecule. Before discussing the oscillations as such, we must considcr the shape and energy scale of the potential energy curve for a molecule of this type. We begin with the knowledge that in a diatomic molecule there is a fairly well defined distance betwecn the nuclei of the two atoms. This is called the bond length and is always of the order of 1 angstrom (10 -10 m). The fact that such an equilibrium distance exists implies that the potential-energy function U(r) has a minimum at the separation r equal to the bond length. One way of defining such a potential, while still leaving plenty of room for empirical adjustment of the constants to experi- mental data, is to assume the following potential-energy function: U(r) = d _ * (10-38) where A and B are positivc constants, and a and b are suitably chosen positive exponents with a > b. The term A/r" is a positive potential energy that falls away rapidly with increasing distance and represents a rcpulsive force. The term - B/r 6 is a negative potential energy, also falling off with increasing r but less rapidly than the first term, and it represents an attractive force. These funetions, and the U(r) curve obtained by adding them, are shown in Fig. 10-19. Now from Eq. (10-38) we can define the equilibrium dis- ■The remainder of this chapter can be omitled without loss of continuity, but it does introduce some quile interesting fcatures in addition to the atomic physics as such. 405 The diatomic molecule Fig. 10-19 Simple model oflhe potential- energy diagram ofa dialomic molecule, defining an eguilib- rium separation ro of the nuclei. tance; it is the distance at which dU/dr = 0: dU dr aA bB f<>+\ ' fi + 1 At r = r , we have aA ro- bB r " (10-39) (10-40) If r o is experimentally known, this defines one connection between the parameters A, B, a, and b. Also, at r = r , the potential energy is given by and substituting for A/r a from Eq. (10-40), this gives ™--£(«-l) The energy U(r ) is a measurable quantity, for if we are to dissociate the molecule completely we must add a quantity of energy, the dissociation energy D, equal and opposite to U(r ) (cf. Fig. 10-19). It is typically a few eV. Thus we can put 406 Energy conservaiion in dynamics 3H) (10-41) Equations (10-40) and (10-41) can be used to narrow down the choice of parameters — we have two equations and four unknowns. Clearly, however, we must either appeal to some further, independent information, or make some specific assump- tions, or both, to have a quantitatively defined situation. The molecular spring constant The quantity we really need to know, for the purpose of con- sidering molecular vibrations, is the effective spring constant of the molecule for small displacements from equilibrium. We dis- cussed this situation in general terms earlier in this chapter (p. 395), and arrived at the result [Eq. (10-29)] t/(s) « \C2S 2 where s = r — ro The constant c 2 can be identified with spring constant k and can be deduced from the given form of U (r): *-(3L-(3)L Applying this to the present problem, we shall differentiate both sides of Eq. (10-39) with respect to r: d 2 U _ a(a+\)A b(b + \)B d r 2 ~ r o+2 r 6+2 UW"«J This Iooks complicated, but for the particular value r = r we can reduce it to one term with the help of Eq. (10-40): (d 2 V\ = a+ 1 bB _ 6(AJ + \)B Therefore, This still looks rather forbidding, but if you compare Eq. (10-44) 407 The diatomic molecule with Eq. (10-41) you will see that the equation for the dissociation energy D contains a very similar combination of factors, and taking D as known we can greatly simplify our statement of the value of k, as follows. From Eq. (10-41) we have (a - b)B ro" = aD Substituting this result in Eq. (10-44) we find k = abD ro 2 (10-45a) Beyond this point we cannot go, even with a knowledge of D and r , without assigning a value to the product ab which can be taken to play the role of a single adjustable parameter, so that we content oursclves with a final semiempirical equation: k = C- r 2 (10-45b) where C is an empirical constant (=ab) greater than unity. Fig. 10-20 Empirical potential-energy diagram for the molecule HCI, based on the Morse potential, with ener- gies expressed in wave numbers (cm~ '). 408 Energy conscrvaiion in dynamics Actually a quite different analytic form of the potential function finds favor with spectroscopists. It is known as the Morse potential (after P. M. Morse). It has the same general features as the one we have used (see Fig. 10-20) but is based on a difference of exponentials rather than of simple powers of r. Like the one we have used, it has adjustable constants that can be deduced from spectroscopic data. Indeed, once the theory of molecular vibrations has been set up, one uses observations to feed back the numerical values into the theoretical formulas. It is a two-way traffic, in other words. Notice the unfamiliar units (cm -1 ) on the ordinate of Fig. 10-20; we shall be coming back to them shortly. The molecular vibrations The period of vibration of a diatomic molecule with atoms of masses /«i and m 2 is given by combining Eqs. (10-36) and (10-45b): ■(&)' T=2xr (-£-) (10-46) where/i = reduced mass = m\m%l(jn\ + m 2 ). The frequency v, in vibrations per second, proves to be a more interesting quantity: -U¥T Let us see what this equation might suggest with an actual mole- cule. We shall take carbon monoxide (CO) for which the follow- ing data apply: /m( l2 C) = 12amu = 2.0 X 10- 2C kg m 2 ( lc O) = 16amu = 2.7 X 10 -26 kg r « 1.1 A- 1.1 X lO-^m 1 D« lOeV = 1.6 X K) -18 ^ First we have _ _m 1 m 3 _ _ 12 x 2? x ]0 _ 20kg _ j 16 x 1() _ 2 o kg /Mi ■+- M»2 ^o 'These values come from a tabulation of molecular constants in the Smithsonian Physical Tables (published by the Smithsonian Institution, Washington, D.C.) 409 Tlic diatomic molecule Therefore, D I y. « 1.4 X 10 8 m 2 /sec 2 whence j- « 1.7 X lO^C'^sec- 1 If we ignored the factor C 1 ' 2 , we would have ► *1.7X 10 13 sec-' Since molecular vibrations are studied mainly through spectros- copy and the measurement of wavelengths of absorbed or emitted radiation, let us calculate the wavelength X corresponding to v: The visible spectrum extends from about 4.5 X 10 -7 m to 7 X 10 -7 m, so the wavelength we have calculated would be well into the infrared. That is exactly where we find the spectral lines associated with molecular vibrations. Actually there is a funda- mental vibrational line of CO with a wavelength of about 4.7 X 10 -6 m— about one quarter of the value we have just calculated. We would have obtained this value by putting C" 2 = 4 in Eq. (10-47). This value of C could correspond, for example, to having a = 5 and b = 3 in the original expression for U(r) [Eq. (10-38)]. This specific form of potential would result from a short-range repulsive force varying as l/r°, and a longer-range attractive force varying as l/r 4 . We should emphasize, however, that our model of the potential is a very crude one and should not be regarded as a source of precise information about intramolecu- lar forces. Our purpose in introducing it is simply to illustrate the general way in which vibrations about an equilibrium con- figuration can be analyzed. There is a feature — a vital feature — that we have ignored; this is that molecular vibration, and the radiation associated with it, is quantized. The energy of a quantum of the radiation is hv (where h is Planck's constant) and is provided by a transition between two sharply defined energy states of the molecule: hv = Ex - E 2 (10-48) We shall not go into this here, except to point out that the smallest possible jump between successive vibrational energy levels does, 410 Energy conservation in dynamics as it happens, correspond exactly to the classical oscillator fre- quency we have calculated. It is the basic quantum nature of the process, as expressed in Eq. (10-48), that makes the frequency v rather than the wavelength X the important quantity. Since, however, spectroscopists measure wavelengths in the first instance, and since v is inversely proportional to X, there is a custom of giving results in terms of X - 1 simply. To convert this to an equiva- lent energy measurement we must use the relation, applying to any photon, that E = hv = hc\~ l This scheme of units has been extended still further, and used in spectroscopy as a measure of energies generally, whether or not photon emission is involved. For example, in Fig. 10-20, the dissociation energy D of HC1 is given as D = 36,300 cm- 1 = 3.63 X 10 6 m" 1 Using the above relation, and substituting h J-sec, we have in this case D = (6.62 X 10" 34 )(3 X 10 8 )(3.63 X 10°) = 7.2 X 10- "J = 4.5 eV (approx.) 6.62 X 10- 34 PROBLEMS 10 — / A particle of mass m, at rest at t = 0, is subjected to a force F whose value at / = is Fq and which decreases linearly with time, becoming zero at t = T. What is the kinetic energy of the particle at t = r? 10-2 An object of mass 5 kg is acted on by a force that varies with the position of the object as shown. If the object starts out from rest at the point x = 0, what is its speed (a) at x = 25 m, and (b) at x = 50 m? 41 1 Problems 10-3 (a) A particle of mass m, initially at rest, is acted upon by a force F which increases linearly with time: F = Cl. Deduce the relationship between F and the particle's position x, and graph your result. (b) How is the graph of F versus x altered if the particle has initial velocity uo? 10-4 (a) Since antiquity man has made use of the "mechanical advantage" of simple machines, defined as the ratio of the load that is to be raised to the applied force that can raise it. By analyzing the work-energy relation dW = F • t/s, and appealing to conservation of energy, give a reasoned definition of "mechanical advantage." (b) In terms of your definition, calculate the particular "me- chanical advantage" offered by each of the devices shown. Radius b 10-5 (a) The Stanford linear accelerator ("SLAC") delivers energy to electrons at the rate of approximately 200 IceV/ft. How does this compare to the energy per unit length imparted to electrons by a cathode-ray tube? By a television tube? (b) The effects of relativity are extremely pronounced in this examplc (after 2 miles of travel, the electrons in SLAC attain a velocity that is within 10 _7 % that of light). How far would the electrons travel in attaining the speed of light if their kinetic energy were given by the classical value KE = £mo 2 ? 10-6 A railroad car is loaded with 20 tons of coal in a time of 2 sec while it travels through a distance of 10 m beneath a hopper from which the coal is dischargcd. 412 Energy conservation in dynamics (a) What average extra force must be applied to the car during this loading process to keep it moving at constant speed? (b) How much work does this force perform? (c) What is the increase in kinetic energy of the coal? (d) Explain the discrepancy between (b) and (c). 10-7 A car is being driven along a straight road at constant speed v. A passenger in the car hurls a ball straight ahead so that it leaves his hand with a speed u relatiue to him. (a) What is the gain of energy of the ball as measured in the reference frame of the car? Of the road? (b) Relate the answers in (a) to the work done by the passenger and by the car. Satisfy yourself that you understand exactly what forces are acting on what objects, over what distances. 10-8 A common device for measuring the power output of an engine at a given rate of revolution is known as an absorption dynamometer. A small friction brake, called a Prony brake, is clamped to the output shaft of the engine (as shown), allowing the shaft to rotate, and is held in position by a spring scale a known distance R away. (a) Derive an expression for the horsepower of the engine in terms of the quantities R, F (the force recordcd at the spring scale), and 03 (the angular velocity of the shaft). (b) You will recognize the product of the force F times its lever arm R as the torque exerted by the engine. Does your relation between torque and horsepower agree with the data published for automobile engines? Explain why there may be discrepancies. 10-9 An electric pump is used to empty a flooded basemcnt which measures 30 by 20 ft and is 15 ft high. In a rainstorm, the basement collected water to a depth of 4 ft. (a) Find the work necessary to pump the water out to ground level. (b) Supposing that the pump was driven by a 1-hp motor with a 50% efficiency, how Iong did the operation take? (c) If the depth of basement below ground werc 50 ft instead of 15 ft, how much work would be required of the pump, again for a 4-ft flood? 413 Problems 10-10 (a) Look up the current world records for shot put, discus, and javelin. The masses of these objects are 7.15, 2.0, and 0.8 kg, respectively. Ignoring air resistance, calculate the minimum possible kinetic energy imparted to each of these objects to achieve these record throws. (b) What force, exertcd over a distance of 2 m, would be re- quired to impart these energies? (c) Do you think that the answers imply that air resistance imposes a serious limitation in any of these events? 10-11 A perverse traveler walks down an ascending escalator, so as to remain always at the same vertical level. Does the motor driving the escalator have to do more work than if the man were not there? Analyze the dynamics of this situation as fully as you can. 10-12 The Great Pyramid of Gizeh when first erected (it has since lost a certain amount of its outermost layer) was about 150 m high and had a square base of edge length 230 m. It is effectively a solid block of stone of density about 2500 kg/m 3 . (a) What is the total gravitational potential energy of the pyramid, taking as zero the potential energy of the stone at ground level? (b) Assume that a slave employed in the construction of the pyramid had a food intake of about 1500Cal/day and that about 10% of this energy was available as useful work. How many man-days would have been required, at a minimum, to construct the pyramid? (The Greek historian Herodotus reported that the job involved 100,000 men and took 20 years. If so, it was not very efficient.) 10-13 It is claimed that a rocket would rise to a greater height if, instead of being ignited at ground level (A), it were ignited at a lower level (B) after it had been allowed to slide from rest down a friction- less chute — see the figure. To analyze this claim, consider a simplified model in which the body of the rocket is represented by a mass M, the fuel is represented by a mass m, and the chemical energy released in the burning of the fuel is represented by a compressed spring be- tween M and m which stores a definite amount of potential energy, U, sufficient to eject m suddenly with a velocity V relative to M. (This 414 Energy corrservation. in dynamics corresponds to instantaneous burning and ejection of all the fuel — i.e., an explosion.) Then proceed as follows: (a) Assuming a value of g independent of height, calculate how high the rocket would rise if fired directly upward from rest at A. (b) Let B be at a distance h vertically lower than A, and suppose that the rocket is fired at B after sliding down the frictionless chute. What is the velocity of the rocket at B just before the spring is re- leased? just after the spring is released? (c) To what height above A will the rocket rise now? Is this higher than the earlier case? By how much? (d) Remembering energy conservation, can you answer a skeptic who claimed that someone had been cheated of some energy? (e) If you are ambitious, consider a more realistic case in which the ejection of the fuel is spread out over some appreciable time. Assume a constant rate of ejection during this time. 10-14 A neutral hydrogen atom falls from rest through 100 m in vacuum. What is the order of magnitude of its kinetic energy in electron oolts at the bottom? (1 eV = 1.6 X 10 -19 J. Avogadro's number = 6 X 10 23 .) 10-15 A spring exerts a restoring force given by F(x) = -kiX- k 2 x* where x is the deviation from its unstretched length. The spring rests on a frictionless surface, and a frictionless block of mass m and initial velocity v hits a spike on the end of the spring and sticks (see the figure). How far does the mass travel, after being impaled, before it comes to rest? (Assume that the mass of the spring is negligible.) 10-16 A particle moves along the x axis. Its potential energy as a function of position is as shown in the figure. Make a careful freehand sketch of the force F(x) as a function of x for this potential-energy curve. Indicate on your graph significant features and relationships. 415 Problcms m, /M2 /0-77 A uniform rope of mass M and length L is draped symmet- rically ovcr a frictionless horizontal peg of radius R. It is thcn dis- turbed slightly and begins to slide off the peg. Find the speed of the rope at the instant it leaves the peg completely. 10-18 Two masses are connected by a massless spring as shown. (a) Find the minimum downward force that must be exerted on m\ such that the entire assembly will barely leave the table when this force is suddenly removed. (b) Consider this problem in the time-reversed situation: Let the assembly bc supported abovc the table by supports attached tonil. Lower the system until m z barely touches the table and then release the supports. How far will m\ drop bcfore coming to a stop? Does knowledge of this distance help you solve the original problem? (c) Now that you have the answer, check it against your in- tuilion by (1) letting mz be zero and (2) letting mi be zero. Espccially in the second case, does the theoretical answer agree with your common sense? If not, discuss possible sources of error. 10-19 A particle moves in a region where the potential energy is given by U(x) = 3x 2 - x 3 (x in m, U in J) (a) Sketch a freehand graph of the potential for both positive and negative values of x. (b) What is the maximum value of the total mechanical energy such that oscillatory motion is possible? (c) In what range(s) of values of x is the force on the particle in the positive x direetion ? 10-20 A highly elastic ball (e.g., a "Superball") is released from rest a distance // above the ground and bounces up and down. With each bounce a fraelion /of its kinetic energy just before the bounce is lost. Estimate the length of time the ball will continue to bounce if h = 5 m and / = rtf. 10-21 The elastic cord of the catapult shown in the diagram has a total relaxed length of 2/ ; its ends are attached to fixed supports a distance 2b apart. 416 Energy conservation in dynamics (a) Show that if the tcnsion developed in the cord is propor- tional to the increase in its length, the component of force in the x direction is F» = 2kb \-}— n - -r-sr J cos \sin 6 sin O / (b) Noting that the position of the stone in the catapult is given by x = —b cot 6, derive the expression for the work done in moving a distance dx. Then integrate this expression between do and 9 to find the total work done in extending the catapult and compare with the result that is obtainable directly by considering the energy stored in the strctched cord (p. 392). (After carrying out the calculation in the way prescribed above, you will better appreciate the advantages that may come from the use of energy conservation instead of a work integral that involves the force explicitly.) 10-22 A perfectly rigid ball of mass M and radius r is dropped upon a deformable floor which exerts a force proportional to the distance of deformation, F = ky. (a) Make a graph of the potential energy of the ball as a func- tion of height y. (Take y = as the undisturbed floor level.) (b) What is the equilibrium position of the center of the ball when the ball is simply resting on the floor? (Note that this corre- sponds to the minimum of the potential-energy curve.) (c) By how much is the period of M increased over its period of bouncing on a perfectly rigid floor? 10-23 A particle of mass m = 2 kg on a frictionless table is at rest at time f = at position x = 0. A force of magnitude F x = 4 sin(ir/) newtons acts on the particle from / = to t = 2 sec. (a) Plot the resulting acceleration a x , velocity o„ and position x as functions of t in this period. (b) What is the total work done by the force during the period t = Otot = 1 sec? 10-24 A spring of negligible mass exerts a restoring force given by F(x) = -kix + k 2 x 2 (a) Calculate the potential energy stored in the spring for a displacement x. Take U = at x = 0. (b) It is found that the stored energy for x = — b is twice the stored energy for x = +b. What is ki in terms of Ari and bt (c) Sketch the potential-energy diagram for the spring as defined in (b). (d) The spring lies on a smooth horizontal surface, with one end fixed. A mass m is attached to the other end and sets out at x = in the positive x direction with kinetic energy equal to k ib' 2 /2. How 417 Problems 'h, \ i. o fast is it moving at x = +6? (e) What are the values of x at the extreme ends of (he range of oscillation? [Use your graph from part (c) for this.] 10-25 An object of mass m, moving from the region of negative x, arrives at the point x = with speed vo. For x > it experiences a force given by F(x) = -ax 2 How far along the +x axis does it get? 10-26 The potential energy of a particle of mass m as a function of its position along the x axis is as shown. (The discontinuous jumps in the value of U are not physically realistic but may be assumed to approximate a real situation.) Calculate the period of one complete oscillation if the particle has a total mechanical energy E equal to 3l/o/2. 2t/„ — i U 10-27 Consider an object of mass m constrained to travel on the x axis (perhaps by a frictionless guide wire or frictionless tracks), atlached to a spring of relaxed length /o and spring constant k which has its other end fixed at x = 0, y ■ / (see the figure). (a) Show that the force exerted on m in the x direction is ~- kx [ l -( l+ $) (b) For small displacements (x « /o), show that the force is proportional to x 8 and hence U{x) ~ Ax* (x « /o) What is A in terms of the above constants? (c) The period of a simple harmonic oscillator is independent of its amplitude. How do you think the period of oscillation of the above motion will depend on the amplitude? (An energy diagram may be helpful.) 10-28 Consider a particle of mass m moving along the x axis in a 418 Energy conservation in dynamics force field for which the potenlial energy of the particle is given by U = Ax 2 + Bx* (A > 0, B > 0). Draw the potential-energy curve and, arguing from the graph, determine something about the de- pendence of the period of oscillation T upon the amplitude xo- Show that, for amplitudes sufficiently small so that Bx A is always very small compared to Ax i , the approximate dependence of period upon ampli- tude is given by ,_ A (i-g„) 10-29 A particle of mass m and energy E is bouncing back and forth between vertical walls as shown, i.e., over a region where U = 0. The potential energy is slightly changed by introducing a tiny rec- tangular hump of height AU (« E) and width Ax. Show that the period of oscillation is changed by approximately (m/2£ 3 ) 1/2 (Ai/AAr). [It is worth noting that the effect of the small irregularity in potential energy depends simply on the product of AU and Ax, not on the individual values of these quantities. This is typical of such small disturbing effects — known technically as perturbations.] U t - U = 10-30 Two blocks of masses m and 2m rest on a frictionless horizontal table. They are connected by a spring of negligible mass, equilibrium length L, and spring constant k. By means of a massless thread connecting the blocks the spring is held compressed at a length L/2. The whole system is moving with speed y in a direction perpendicular to the length of the spring. The thread is then burned. In terms of m, L, k, and o find (a) The total mechanical energy of the system. (b) The speed of the center of mass. (c) The maximum relative speed of the two blocks. (d) The period of vibration of the system. How do the quantities of parts (a) through (d) change if the initial velocity c is along, rather than perpendicular to, the length of the spring? 10-31 The mutual potential energy of a Li + ion and an I - ion as a function of their separation r is expressed fairly well by the equation 419 Problems -Ke 2 . A £/(,) = __ + __ where the first term arises from the Coulomb interaction, and the values of its constants in MKS units are K = 9 X 10 9 N-m 2 /C 2 , e = 1.6 X KT 19 C The equilibrium distance ro between the centers of these ions in the Lil molecule is about 2.4 A. On the basis of this information, (a) How much work (in eV) must be done to tear these ions completely away from each other? (b) Taking the I - ion to be fixed (because it is so massive), what is the frequency v (i n Hz) of the Li + ion in vibrations of very small amplitude? (Calculate the effective spring constant k as the value of d 2 U/dr 2 at r = ro — see p. 407. Take the mass of the Li + as 10- 26 kg.) 10-32 (a) If in addition to the van der Waals attractive force, which varies as r -7 , two identical atoms of mass M experience a repulsive force proportional to r - ' with / > 7, show that "W-7T + I <" >6 > and graph your result versus r. (b) Calculate the equilibrium separation ro in this molecule in terms of the constants by requiring dU(r) dr = (c) The dissociation energy D of the molecule should be equal to — t/(r ). What is its value in terms of A, n, and r ? (d) Calculate the frequency of small vibrations of the molecule about the equilibrium separation r . Show that it is given by the mass M, the constant n, the equilibrium separation, and the dissociation energy of the molecule, as follows: „ \2nD 10-33 The potential energy of an ion in a crystal lattice of alternatcly charged ions may be written where A, B, and n are constants, e is the elementary charge, and r is the distance between closest neighbors in the lattice, called the lattice constant. This potential arises from the l/r 2 force of electrostatic 420 Energy conservation in dynamics attraction, together with a short-range repulsive force as two adjacent ions are brought close together. (a) Make a graph of U(r) versus r. Making sure to identify r correctly, what can you say about the stability of the crystal structure? (b) Calculate the lattice constant ro in terms of A, B, e, and n, from the equilibrium condition dU(r) dr (This defines the inter-ionic distance for which the energy of the lattice as a whole is minimized.) (c) Show that the binding energy per mole of molecules is u = —NolKro) N o = Avogadro's number ~6 X 10 23 mole -1 Using the result of part (b), determine u in terms of r . (d) For NaCl crystal, the equilibrium lattice spacing ro is 2.8 A, and the exponent n is about 10. The constant A is equal to 1.7 (dimensionless) if r is expressed in centimeters and e is given in esu (e = 4.8 X 10 -10 esu). Using these values, calculate a theoretical value for u and compare with the experimental value of 183 kcal/mole. 421 Problems A technique succeeds in mathematical physics, not by a clever trick, or a happy accident, but because it expresses some aspect ofa physical truth. o. G. sutton, Mathematics in Action (1954) 11 Conservative forces and motion in space EXTENDING THE CONCEPT OF CONSERVATIVE FORCES throughout chapter 10 we consistently applied one important simplification or restriction, by confining our discussion to motion in one dimension only. This clearly prevented us from studying some of the most interesting and important problems i n dynamics. In the course of the present chapter we shall free ourselves of this restriction and in the process show the energy method of analysis to still greater advantage. To begin the discussion, let us consider a problem in motion under gravity near the earth's surface. Suppose we have two very smooth tubes connecting two points, A and B, at different levels in the same vertical plane (Fig. 11-1). A small particle, placed in either of these tubes and released from rest at A, slides down and emerges at B. If the tubes are effectively frictionless, the forces exerted by them on the particle are always at right angles to the particle's motion. Hence these forces do no work Path2 Fig. 11-1 Alternatice paths between Path 1 two gtoen points f or motion in a vertical plane. 423 on the particle ; whatever changes may occur in its kinetic energy cannot be ascribed to them. What these forces do achieve is to compel the particle to follow a particular path so that it emerges at B traveling in a designated direction. If it follows path 1, it emerges with a velocity v t as shown; if it follows path 2, it emerges with a velocity v 2 . Of course the energy of the particle does change as it moves along either tube; the gravitational force is doing work on it. But we observe a very interesting fact: Although the directions of the velocities Vi and v 2 are quite different, their magnitudes are the same. The kinetic energy given to the particle by the gravitational force is the same for all paths beginning at A and ending at B. How does this come about? It is not difhcult to see. As the particle travels through some clement of displacement ds the work dW done on it is given by dW = |F„| \ds\ cos 6 = F„ • ds where B is the angle between the directions of F„ and ds. But the force F„ is a constant force (i.e., the same at any position) with the following components: Fx = F„= -mg (we take the y axis as positive upward). The element of dis- placement has components (dx, dy). Now from the basic defini- tion of dW as |F| \ds\ cos 6 we have dW = F x dx + F„ dy To see this, consider any two vectors A and B in the xy plane, making angles a and /3, respectively, with the x axis (Fig. 11-2). Then if the angle between them is 6 we have, by a Standard trigonometric theorem, that cos 6 = cos(/3 — a) = cos /3 cos a + sin/3sina The scalar product 5 (= A • B) is thus given by S = |A| |B| (cos /3 cos a + sin/3sina) Fig. 11-2 Basis for obtainiiiR the scalar product A • B in terms of individual components, as may be convenient in calculaling the work F • ds in an arbilrary displace- ment. 424 Conscrvative forces and molion in space But |A| cosa = A x |A| sina = A„ and similarly for the components of B. Thus we have A • B = A X B X + A u By (two dimensions only) More generally, if the vectors also have nonzero z components, A • B = A X B X + A y B u + A,B, Thus in general we shall have dW = F • ds = F x dx + Fydy + F, dz (11-1) In the present two-dimensional problem, with F x = and F y = —mg, this gives us dW = —mgdy Hence for a change of vertical coordinate from y\ to y 2, regard- less of the change of x coordinate or of the particular path taken, we have a change of kinetic energy given by — /. K 2 -K l = JdW= -mg(y 2 - yi ) (11-2) which exactly reproduces the result that we derived in Chapter 10 for purely vertical motion and which permits us again to define a gravitational potential energy U(y) equal to mgy. This result makes for a great extension of our energy methods to situations wherc, in addition to gravity, we have so-called "forces of constraint," which control the path of an object but, because they act always at right angles to its motion, do nothing to change its energy. Let us consider some specific examples. ACCELERATION OF TWO CONNECTED MASSES We shall begin with a problem which, if tacklcd by the dircct application of Newton's laws, requires us to write a statement of F = wa for each of two objects separately. The problem is to find the magnitude of the acceleration of two connected masses, mi and m 2 , moving on smooth planes as shown in Fig. 11-3. The accelerations of the masses are different vectors, a! and a 2 , 425 Acceleration of two connected masses Fig. 1 1-3 Motion ofiwo coimected masses — a simple example of the use of energy-conseruation methods. even though they have the same magnitude a. Thus a single statement of F = ma will not suffice. Using the scalar quantity energy, however, we can exploit the fact that the magnitudes of the displacements and their time derivatives are common to both masses. Thus we have |vi| = [v 2 | = v, so that the total kinetic energy of the system is given simply by K= J(mi +m 2 )v 2 and its change in a time dt is given by dK = («u + mi)odv = {m\ + mi)oadt Now u dt is the distance moved by each mass parallel to the surface on which it rides, and for m 2 the distance v dt down the slope means a change dy in vertical coordinate equal to —v dt sin d (y positive upward), The associated change dU in gravitational potential energy is thus dU = -m2gvdtsin6 But, given conservation of total mechanical energy, assuming friction to be absent, we have dK + dU - Thus (mi + m£)va dt — m2go dt sin 6 = whence the familiar result rri2 m\ + /712 gsintf OBJECT MOVING IN A VERTICAL CIRCLE Suppose that a particle P of mass m is attached to one end of a rod of negligible mass and length /, the other end of which 426 Conservative forces and motion in space / / I I \ \ \ s Fig. 11-4 Mo t ion ofa simple pendulum. ^- ^ ' r "~ a y = O is pivoted freely at a fixed center C. Let us take an origin O at the normal resting position of the object (Fig. 1 1-4). Then the position of the object is conveniently described in terms of the single angular coordinate B, or, if we prefer, by the displacement 5 along the circular arc of its path from O (assuming the rod to be of invariable length). ' If the angular displacement is B, the y coordinate of the object is given by y = /(1 - cos 0) and hence its potential energy by 1/(0) = mgl(\ - cos 8) (11-3) Using our basic energy-conservation statement we have Ki + t/i = K 2 + U 2 for any two points on the path. Substituting for K in terms of m and v and for U by Eq. (11-3), we have \mui 2 + mgl(l - cosfli) = \mv 2 2 + mgl(\ - cos 02) Therefore, V2 2 = oi 2 + 2gl(cosB 2 - cos 0,) (11-4) Clearly, if 8 2 > 6 U then v 2 < V\, If the object were started out at the lowest point with a velocity v , we should have ■In a sense, therefore, the motion is one-dimensional, even though the one dimension is not a straight line. But we shall not stress this aspect of such a system at this point. 427 Object moving in a vertical circle 02* = P0* - 2^/(1 -COS0 2 ) With the help of this result we can answer such questions as: What initial velocity is needed for the object to reach the top of the circle? and: What position does it reach if started out with less than this velocity? Notice the great advantage that the energy method has over the direct use of F = ma in this problem. The velocity of the object is changing in both magnitude and direction, it has radial and transverse components of acceleration, there is an unknown push or puli on the object from the rod — yet none of these things need be considered in calculating the speed v at any given (or y). Once v has been found by Eq. (1 1^) [or perhaps by going back to Eq. (11-2) if that is more convenient], then one can proceed to deduce the acceleration components and other things. There may be subtleties in such problems, however. Sup- pose, for example, that instead of a more or less rigid rod we had a string to constrain the object. This is now a constraint that works only one way; it can puli radially inward but it cannot push radially outward. One may have a situation in which the velocity is not great enough to take the object to the top of the Fig. Jl-5 (a) Path o f the bob of a simple pendulum launched with insufficient velocity to maintain a tension in the string Up to = w. (b) Stroboscopie photograph of a bati falling away from a circular channel at the point where the contact force becomes zero. (Photograph by Jon Rosenfeld, Edueation Research Cenler, M.l.T.) (a) (b) 428 Conservalive l'orccs and molion in spacc circle (although it might be possible with a rigid rod); the object will fail away in a parabolic path [Fig. 1 l-5(a)]. The breakaway point is reached when the tension in the string just falls to zero and the component of F„ along the radius of the circle is just equal to the mass m times the requisite centripetal acceleration v 2 /L An exactly similar situation can arise if an object moves along a circular track made of grooved metal, and Fig. ll-5(b) shows a stroboscopic photograph of an object falling away from such a track at the point where the normal reaction supplied by the track has fallen to zero. The angular position, m , at which this occurs [cf . Fig. 1 l-5(a)] is defined by a statement of Newton's law : < r mg cos(tt - <L) = ' 2 nv 1 ( assumes v <e, where u* = vo 2 - 2*/(l - ■ cos 0„) This leads to the result COS0 m = 2 oo 2 3 3gl We can thus deduce that a particle that starts out from O with o less than VJgl will fail to reach the top of the circle, whereas with a rigid rod to support it an initial speed of 2\/gl would suffice. Notice, therefore, that the energy-conservation principle should not be used blindly; one must always be on the alert as to whether Newton's law can be satisfied at every stage with the particular constraining forces available. AN EXPERIMENT BY GALILEO It was Galileo, in his Dialogues Concerning Two New Sciences (1638), who first clearly stated the result that the speed attained by an object descending under gravity depends only on the vertical distance traveled. He applied this result to the uniformly accelerated motion of an object down smooth plancs of different slopes, as indicated in Fig. 1 l-6(a), which is based on a diagram from Galileo's own book. He could not, however, directly demonstrate the corrcctness of this proposition using inclined planes as such; instead, he performed a very clever experiment 429 An experiment by Galileo Fig. 11-6 (a) Speed atlained by a block sliding down a frictionless plane depends only on the verlica! disiance traveled, not on the slope. (b) Galileo's pendulum ex- periment to demonstrate the properties ofmotion on idealized inclined planes. with a mass swinging in a circular arc, and applied his remarkable scientific insight to what he observed in this situation. Here is an account of the experiment, taken with only minor changes of notation from Galileo's own description; the clarity and modernity of his presentation is quite striking: Imagine this page to represent a vertical wali, with a nail driven into it; and from the nail let there be suspended a lead bullet of one or two ounces by means of a fine vertical thread, OB, say from four to six feet long [see Fig. ll-6(b), based on Galileo's own sketch]. On this wali draw a horizontal line AA', at right- angles to the vertical thread OB (which hangs about two finger- breadths in front of the wali). Now bring the thread OB with the attached ball into position OA and set it free; first it will be observed to descend along the arc ABA' . . . till it almost reaches the horizontal AA', a slight shortage being caused by the re- sistance of the air and the string. From this we may rightly infer that the ball in its dcscent through the arc AB acquired a momentum on reaching B which was just sufficient to carry it through a similar arc BA' to the same height. Having repeated this cxperiment many times, let us now drive a nail into the wali close to the perpendicular OB, say at N, so that it projects out some five or six finger-breadths in order that the thread, again carrying the bullet through the arc AB, may strike upon the nail N when the bullet reaches B, and thus compel it to traverse the arc BC, described about N as 430 Conservative forccs and motion in space center . . . Now, gentlemen, you will observe with pleasure that thc ball swings to the point C in the horizontal line AA', and you would see the same thing happen if the obstacle were placed at some lower point. . . . ' In this way Galileo convinced himself that the speed ac- quired by an object in descending any vertical distance is suf- ficient to carry it up through an equal vertical distance by a different path. He then added to this the reversibility of the motion along any given arc and could reach his main conclusion that the speed attained by his suspended object must be the same whether it descends along the arc A B or the arc C B, or any other arc beginning on the level A A' and with its lowest point at B. Finally, he could visualize such motion as taking place on a continuous succession of inclined planes of different slopes, and so formulate his proposition about uniformly accelerated motion along different inclines. This was an enormously important result, because he was then able to make inferences about free fail under gravity by observing the motion of an object rolling down an inclined plane with a very gentle slope. This per- mitted him to stretch thc time scale of the whole phenomenon (a = gsin 6!) and make quantitative measurements of distance against time, using as his clock a watcr-filled container with a hole in it that he could open and close with his finger— a bril- liantly simple device. MASS ON A PARABOLIC TRACK Suppose we bend a piece of very smooth metal track so that it has the shape of a parabola curving symmetrically upward with its vertex at point O (Fig. 1 1-7). Let the equation of the parabola be y = JCx 2 (11-5) where C is a constant. Imagine an object of mass m, free to slide along the track with negligible friction. Its gravitational potential energy, re- ferred to an arbitrary zero at O, is given by U(y) = mgy ■Quoted from the English translation of Dialogues Concerning Two New Sciences (H. Crcw and A. de Salvio, translators), Dover l'ublications, New York. 431 Mass on a parabolic track Fig. 11-7 Motion of a particle on a parabolic track in a vertical plane. which, by Eq. (1 1-5), we can alternatively write as a function of x: U(x) = \mgCx 2 = \kx 2 (11-6) where k is a constant of the same dimensions (newtons per meter) as a spring constant. At any point {x, y) the mass has a speed v, necessarily along the track, and by conservation of energy we have \mv 2 + U = E (= const.) Denoting the x and y components of v by v x and u„, and using Eq. (11-6), we have %mv x 2 + Imv,* + ikx 2 = E (11-7) This has a marked resemblance to the energy-conservation equation of the linear harmonic oscillator. Only the presence of the term \mo y 2 spoils things. This suggests to us that if we had a situation in which v y were very small compared to v x , the motion of the mass on this track would closely approximate that of a harmonic oscillator. What do we need to achieve this? Common sense tells us more or less immediately that if the track is only very slightly curved, so that y « x at each point, the vertical motion is very small compared to the horizontal and approximately harmonic motion will occur. Such motion can be achieved and beautifully demonstrated with a "linear air track" 1 in which a metal glider rides on a cushion of air blown up through holes in the track. Because of the low friction, the oscillation can take place even if the curvature is extremely small. To take an example, here is a specification of the shape of the track in an actual demonstration (the setting was by machine screws at 1-ft spacings; hence the lapse from MKS): 'R. B. Runk, J. L. Slull, and O. L. Anderson, Am. J. Phys., 31, 915 (1963). 432 Conservativc forces and motion in space X, ft ±1 ±2 ±3 ±4 ±5 ±6 ±7 y,in. & A A 1 4 ff & M Taking any pair of these values, one establishes the value of the constant C in Eq. (11-5), e.g., x = 4ft = 1.22 m y - i in. = 6.35 X 10"» m Thus C = 2y/x 2 = 8.55 X K)- 3 m" 1 What would be the cxpected period of the motion? For mass m and "spring" constant k, we have T = 2r(m/ky 12 , and, from Eq. (11-6), k = mgC. Therefore, m/ k = (gC)"' « 11.9sec 2 7"- 2»vTT9« 21.6 sec The measured period was within a few tenths of a second of this and independent of amplitude to the accuracy of the mea- surement. In this example (and the others of motion under gravity) the moving object, i n its two-dimcnsional motion, really does ride a contour that corresponds in form to the graph of U(x) against x. These are truc two-dimensional motions; they arc physically distinct from the onc-dimensional situations discussed in Chapter 10. To appreciate the difference between them, consider two quite realizable physical systems. The first one [Fig. ll-8(a)] is a mass m attached to a horizontal spring and resting on a smooth table. The second [Fig. ll-8(b)] is an equal Fig. 11-8 (o) Mass attached to a horizon- tal spring. (b) Same mass on a smooth parabolic track. (c) Arrangements (a) and (b) can be made to have identical varia- tions ofpotential energy with x, bui the motions are different. 433 Mass on a parabolic track mass m free to slide on a parabolic track. The strength of the spring and the curvature of the track are adjusted so that the systems have the same parabolic potential energy curve of U(x) against x, as shown in Fig. ll-8(c). If projccted horizontally from the origin with the same speed v , both masses will have the same magnitude of velocity at any other value of x. If set into oscillation, both systems will be periodic. But their periods will be different. The mass on the spring will have the same period at all amplitudes ; the other system will show a progressive change of period with amplitude (which way do you think it will be?). It is only in special situations, such as that with the very slightly curved track, that the two motions become essentially the same. THE SIMPLE PENDULUM In diseussing the simple pendulum, we are returning to the problem of an object moving in a vertical circle. This time, howevcr, we shall go into it more carefully. It is an important type of physical system, over and above its use in eloeks, and it is not quite so simple as its traditional name implies. The sim- plicity is primarily in its structure — idealized as a point mass on a massless, rigid rod. As we saw earlier [Eq. (11-3)], the potential energy, U(6), expressed in terms of the angle that the supporting rod makes with the vertical, can be written in the form U = mgl(l - cos 0) = 2mg/sin 2 I "- (11-8) Ifwe plot this expression for U as a funetion of d, we get Fig. 11-9. Fig. 11-9 Energy diagram for a rigid pendulum, using the angle 6 as the coor- dinate. The pendulum is trapped in oseil/a- tory motion about 6 = Oi/Eis/ess than 2mgl. 434 Conservative forees and motion i n space In Fig. 1 1-9, 6 can take all values from — oo to + cc . How- ever, all positions of the pendulum in space are described by values of 6 between —t and -\-k. Any value of d outside this range corresponds to one and only one angle 6 inside the range. The latter is obtained by adding to or subtracting from the former a whole-number multiple of 2r. Two kinds of motion are possible, depending on whether E is less than or greater than 2mgl (referred to a zero of energy defined by an object at rest at the lowest position of the pen- dulum bob). If the total energy is sufficiently great (e.g., as for E 2 in the figure), there are no turning points in the motion; 8 increases (or decreases) without limit, corresponding to continucd rotation of the pendulum rather than oscillation. The speed of the pen- dulum bob is maximum at = (or ±2ir, ±4ir, etc.) and minimum at 6 = ±tt (or ±3tt, ±5ir, etc.). It is clear from Fig. 11-9 that to produce such rotational motion the pendulum must have a kinetic energy at least equal to 2mgl at the lowest point of the swing. If E < 2mgl, say E y, the motion is oscillatory and the angle B changes from +6q to — 6q and back again (Fig. 11-9). However, the motion is not harmonic except for sufficiently small amplitudes. In the neighborhood of O, one can very nearly match the potential-encrgy curve of Fig. 11-9 by a parabola (Fig. 11-10), so that for these small amplitudes the oscillations are just those of the linear oscillator. One way of justifying this statement is to recall the general argument, made in Chapter 10, that almost any symmetrical potential-energy curve can be approximated by a parabola over some limited range of small displacements. Another way is to note that in Eq. (1 1-8), if d Fig. 11-10 Potential- energy diagram of a simple pendulum. 435 The simple pendulum is small, one can set sin(0/2) approximately equal to 0/2, so that we have U(8) « Imgie 1 (11-9) A third way depends upon using the same approximation that Newton used when he described the moon as a falling object. Applying it to the present problem, and taking the origin at O, the circular path is given by the equation x 2 + (l - y) 2 = / a (seeFig. 11-11). Therefore, y 2 - 2ly + x 2 = y = /-(/ 2 -* 2 )" 2 = / ->('-i) 2\l/2 If x 2 <5C l 2 , we can use the binomial expansion to yield the approximate result 1 £ 2 l (11-10) [Compare this with Eq. (11-5); you will then realize what the constant C in that equation represents. In the actual experiment described, the motion was like that of a simple pendulum nearly 400ftlong!] Equation (11-10) suggests a somewhat different approxima- tion for small oscillations of the pendulum, describing its motion in terms of its horizontal displacement instead of its angular one. But whatever analysis we adopt, it is clear that the period must depend on the amplitude. We can also argue qualitatively which way it varies. Looking at Fig. 11-10, we can say that the curve Fig. 11-11 Geomelry ofilisplace- ments of a simple pendulum. l-y 436 Conservalive forccs and motion in spacc of U(6) for the pendulum describes a kind of spring that gets "softer" at large extensions. Compared to the parabolic behavior that would hold for an ideal spring, the restoring force is rela- tively less at larger displacements. Thus one might guess that the motion becomes more sluggish and hence that the period increases. Certainly there is one extreme case that leaves" no room for doubt. If the energy is such that the pendulum just exactly arrives at n = jr, there is no restoring force at all; the pendulum would sit upside down indefinitely — although in practice this is of course an unstable equilibrium. Even at amplitudes short of this the increase of period is drastic. If you are interested, the section after next describes the analysis of larger-amplitude motion. But first we shall make a detailed study of the small-angle approximation. THE PENDULUM AS A HARMONIC OSCILLATOR The speed of the pendulum bob at any point is equal to / dd/dt, so that the cncrgy-conservation cquation is '(D' 5^/ 2 (^) + U{6) = E (exact) Using the approximate expression for U(6) from Eq. (11-9), we have iml 2 {^) + $mgl0 2 = E (approx.) or (f) Tt) + ]° 2 ° const - (n-u) By now we have met this form of equation scveral times and can identify (g/l) 1 ' 2 as being the quantity w that defines the period of the oscillation: -2r-*(i)' /2 (11-12) [Caution: It is the angular displacement itself that undergoes a simple harmonic variation, described by the equation 0(0 = O sin(oo/ 4- v,,) (11-13) 437 The pendulurrms a harmonic oseiliator This can be confusing; the actual angular displaccment of the pendulum is deseribed in terms of the sine of the purely mathe- matical phase angle (w/ + <p ), The actual angular velocity of the pendulum is dB/dt— not the w in (ut + <p ), which serves merely to define the periodicity.] The behavior of the simple pendulum as a elose approxima- tion to a harmonic oscillator is so important that we shall com- ment on it further. First, let us take a moment to consider the derivation of its equation of motion by a direct application of Newton's law. We can do this in two different ways, according to whether we analyze the problem in terms of linear or angular displacements. Linear motion We consider the horizontal foree aeting on the pendulum bob when it is a horizontal distance x from its equilibrium position [Fig. 1 1— 12(a)]. If the tension in the suspending string is F T , we have 2 m— = -F r sin0 = -F T -. at* I If the vertical component of acceleration of the bob is negligibly small, we can, however, put Ft cos 6 « F a = mg For small angles 6, cos 8 ~ 1 - d 2 /2 w I, SO that F T « mg and the equation of horizontal motion becomes Fig. 11-12 Basis for analyzing the motion of a simple pendulum (a) in terms of hori- zontal displacements, and (b) in terms of angular displacements. 438 Conservalive forees and motion in space J 2 d x mg m W~~T x leading once again to Eq. (11-12) for the period T. Angular motion In this case we consider the transverse acceleration a e (= l d 2 8/dt 2 ), which is produced by the component of F„ in the direction of the tangent to the circular arc along which the pen- dulum bob moves (see Fig. ll-12b). The magnitude of the tension force F^ (which actually varies with time) does not enter into this treatment of the problem. We have, simply, rf 2 w/—- = — F„sin0 = — mg s'm 6 For small 0, sin « 9, and so we have which leads to Eq. (1 1-13) for as a function of time. It can be seen that the analysis in terms of the angular variables is a "cleaner" treatment than the other, involving only the one ap- proximation sin 9 « 9. The isochronous behavior of the simple pendulum — the fact that its period is almost completely independent of amplitude over a wide range — provides a striking example of this remark- able and unique property of systems governed by restoring forces proportional to the displacement from equilibrium. Sup- pose we had two identical pendulums, each with a string of length 30 ft, suspended from a high support. Each would have a period of about 6 sec. And if we set one swinging with an amplitude of only an inch or two, so that its motion was almost imperceptible, and set the other swinging with an amplitude of 5 ft, so that it swept through the central position at a speed of about 5 ft/sec, the difference in their periods would be too small to put them significantly out of step in less than a hundred swings or so. This same isochronous property of the pendulum was of great and perhaps crucial help to Newton, Huygens, et al. when they made their fundamental experiments on collision phenomena (cf. Chapter 9). AH these experiments were done with masses suspended from equal strings. As long as the small-amplitude 439 The pendulum as a liarmonic oscillatoi (b) Fig. 11-13 (a) Two pendulums ofeqnal Icngth are re- leasedfrom arbitrary posltions al ihe same instan/, (b) The isochronous properly of ihe pendulum ensures that ihe collision occurs w/ien bolh masses are al Ihe boltom politis of iheir swing. approximation is valid, two masses released at the same instant from arbitrary positions [Fig. ll-13(a)] will rcach thcir lowcst points at thc same instant [Fig. ll-13(b)], so that the collision occurs when each mass is traveling horizontally with its maximum vclocity— thc magnitude of which is given by the equations of simplc harmonic motion. ' Even for angular amplitudes that are large enough to require the use of the exact equation for p m „ , t'Lx = 2gl{\ - cos O ) the colliding masses will reach their lowest points at almost the same instant as long as O is less than 90°, which it would have to be for an object attached to a string rather than to a rigid rod. The next section briefly discusses the detailed dependence of period on amplitude for a simple pendulum. THE PENDULUM WITH LARGER AMPLITUDE 2 To find how the period of a simple pendulum departs from its ideal small-amplitude value, we write the equation for con- servation of energy in the exact form ©' + 2<oo 2 (l - cos 0) = 2coo 2 (l - cos 8 ) (11-14) where w,, 2 = g/l and #o > s tne maximum angle of dcflcction from the vertical. The period of oscillation is thcn given by 7X0,,) = wo" dO fs/2 J-t (co$8 - cos<?o) 1/2 (11-15) 'You can easily vcrify that the magnitude of this velocity is proportional, in thc small-amplitude approximation, to the horizontal distance through which a mass is initially drawn aside before bcing released. This makes analysis of thc cxperiments very simple. 2 This section may be omitted without loss of continuity. 440 Conservative forces and motion in space Fig. 11-14 Precise measuremenls on the period of a simple pendulum, showing the quadratic increase of period wilh amplitude, and the close approach to isochronism ob- tained by damping the top endofthe supporting wire be- tween cycloidal jaws, so that the free length shortens as the ampli- tude increases. For small amplitudes, we of course have a period T equal to 2w/(j)o, as discussed iri the previous section. The integral of Eq. (11-15) cannot be carried out exactly; one has to resort to numerical methods or to a series expansion of the integrand which gives, as a next approximation to the period, the following result: 7tyo)« To 1 +£sin ['+**>'(?)] (11-16) If O is not too large, another acceptable form of this result, in terms of the horizontal amplitude A (= /sin 6 ) is the formula T(A) « To (■+tf) (11-17) Figure 11-14 shows the results of some precision measurements that verify Eq. (11-17). 1 The pendulum had a length of about 3 m, and the greatest amplitude used was about 0.5 m, so that values of 6 up to about 10° (« 0.17 rad) are represented. It is worth noting that the over-all change of T from the least to the greatest values of d studied is less than 2 parts in 1000; the in- dividual points are accurate to about 1 part in 10 5 , and the validity of Eq. (11-17) is very nicely demonstrated over this range of O - The graph also shows the results of using a special 'M. K. Smith, Am. J. Pftys., 32, 632 (1964). 441 The pendulum with larger amplitude pendulum support that shortens the pendulum as 6 increases, in such a way that the tendency for T to increase with O >s almost completely compensated. Ideally, the support should be shaped so that the path of the pendulum is a portion of a cycloid curve. (Incidentally, the method of making an exactly isochronous pendulum, through the use of a cycloidal suspension, was first discovered by Huygens. In 1673 he published a great book, the HoroJogium Oscillatoriwn, in which many important dynamical results were first presented within the framework of the very practical problems of clock design.) UNIVERSAL GRAVITATION: A CONSERVATIVE CENTRAL FORCE The problems of energy conservation that we have discussed so far have involved only the familiar force of gravity near the earth's surface— a force which, as experienced by any given object, has effectively the same magnitude and the same direction in a given locality, regardless of horizontal or vertical displace- ments (within wide limits). 1 But, as we know, the basic gravita- tional interaction is a force varying as the inverse square of the distance between two particles and exerted along the line joining their centers. It is an example of the vcry important class of forces— central forces— which are purcly radial with respect to a given point, the "center of force". It has the further property of being sphcrically symmetric; that is, the magnitude of the force depends only on the radial distance from the center of force, and not on the direction. We shall show that all such spherically symmetric central forces are conservative, and shall then con- sider the special features of the l/r 2 force that holds for gravita- tion and electrostaties. If a partiele is exposed to a central force, the force veetor F aeting on it has only one component, F T . If the force is also spherically symmetric, then F r can be written as a funetion of r only: F r =/(r) (11-18) We shall prove that any such sphcrically symmetric central force is conservative by showing that the work done by the force on a •Recall, howcvcr, that small local variations are in Tacl detectable by sensitive gravity survey instruments, as deseribed in Chapter 8. 442 Conservative forces and motion in space __M— _______ ___■) (a) 7%. 77-75 (a) Diagram for eonsideration ofpotential- energy changes in a Central force field. (b) Analysis of an arbitrary pai h into radial elements along which work is done, and transverse elements along which no work is done. (c) Closed path in a conservative central force field. test particle, as the latter changes its position from a point A to another point B [Fig. II-I5(a)], does not depend on the par- ticular path connecting A and 73, but only on these end points. If this result holds, then it will be true that the particle, if it went from A to B by this path, and returned from B to A by any other path, would have no net work done on it by the central force and would (if no work were done on it by other forces) return to A with the same kinetic energy that it had to start with. This then corresponds exactly to the conservative property as defined for one-dimensional motions. Let the center of force be at O [Fig. ll-15(a)], and consider the work done by the central force F on the test particle as it undergoes a displacement ds along the path as shown. This work is given by dW = F • ds = F ds cos a (11-19) where a is the angle between the direction of F, i.e., the direction of r, and the direction of ds. From the figure, however, we see that ds cos a = dr where dr is the change of distance from O, resulting from the displacement ds. Inserting this value of ds cos a into Eq. (1 1-19), we have dW = Fdr 443 Universal gravitation: a conservative central force where the magnitude of the force [F = /(/■)] depends only on r as indicated in Eq. (1 1-18). We then have for the work done by F on the test object as it moves from A to B, r 'b W= / f(r)dr (11-20) •> 'A Because this integral has a value that depends only on its limits and not on the path, we can conclude that the spherically sym- metric central force is conservative. This result can be arrived at in slightly different terms by picturing the actual path as being built up from a succession of small steps, as shown in Fig. ll-15(b). One component of each step is motion along an arc at constant r, so that the force is perpendicular to the displace- ment and the contribution to W is zero, and the other component is a purely radial displacement so that the force and the dis- placement are in the same direction, resulting in the amount of work F dr by the central force. [A converse to the result we have just derived is the following important proposition : A central force fleld that is also conservative must be spherically symmetric. To show this, suppose that a center of force exists at the point O in Fig. ll-15(c). Imagine a closed path ABCD, formed by very short portions of two radial lines drawn from O, and by the two circular arcs BC and DA. Since, by definition, the force is purely central, it has no component perpendicular to the radial direction from O at any point. Thus, if we imagine a particle carried around ABCD, it cxperiences no force along BC and D A. The condition that the force be conservative thus requires equal and opposite amounts of work along AB and CD. Since these lines are of equal length, the mean magnitude of the force must be the same on each. If we imagine the lengths of these elements of path to become arbitrarily small, we conclude that the value of the force at a particular scalar value of r is independent of the direction in which the vector r is drawn. The most important sources of such spherically symmetric force fields are spherically symmetric distributions of mass or electric charge.] It is important to realize that a force may be conservative without being central or spherically symmetric. For example, the combined gravitational effcct of a pair of concentrated un- equal masses, separated by some distance, has a complicated 444 Conservative forces and motion in space dependence on position and direction, but we know that it is conservative because it is the superposition of two individually conservative force fields of thc separate masses. Given the rcsult expressed by Eq. (11-20), we can proceed to define a potential energy U(r) for any object exposed to a spherically symmetric central force: r 'd = - / f (r) J 'A V B - U a = - / f(r)dr (11-21) If the kinetic energy of the object at points A and B has the values K A and K B , respectivcly, then (if no work is done by other forces) we have Kg - Ka = W and Ka + U A - K B + Ub - B (11-22) Thus we have established an energy-conservation statement for an object moving under the action of any central force. For an inverse-square force, we have F(r) = ^ 2 dl-23) In such a case, therefore, Eq. (11-21) gives us r 'n U B - U a C r" 2 dr J 'A from which we get . Vb r a) Ub- Ua- Cl — -— J (11-24) Vb r a) There remains only the choice of the zero of potential energy. It is convenient to set U = for r = ao, i.e., at points infinitely far from the source at O, since thc force vanishes at these points, and in colloquial terms one can say that the existencc of either particle is of no consequence to the other one under these con- ditions. We now apply Eq. (11-24) to the case where r A = y-, U A = 0, and the potential energy of thc test particle becomcs Ub = C/rji — or, if we drop the now redundant subscript B, there follows 445 Universal gravitation; a conservative central force V(r) = - f (11-25) for the potential energy of a test particle as a function of its position. Equation (11-25) is valid for either an attractive or a re- pulsive force, the constant C being negative for attractive forces and positive for repulsive forces. In particular, for a particle of mass m under the gravitational attraction of a point mass M (which we shall suppose to be fixed at the origin O) we have F(r)=-^ (.1-26) DW--2? 01-27) Note that these last two equations refer only to the inter- action between two objects that can be regarded as though they were points — i.e., their linear dimensions are small compared to their separation. As we saw in Chapter 8, however, some of the most interesting and important gravitational problems concern the gravitational forces exerted by large spherical objects such as the earth or the sun. In our earlier discussion of such prob- lems, we saw that the basic problem is the interaction between a point particle and a thin spherical shell of material. In Chapter 8 we presented a frontal attack on this problem, going directly to an evaluation of an integral over all contributions to*the net force. Now we shall approach the problem by way of a con- sideration of potential energies, and in the process we shall see the great value of the potential energy concept in such cal- culations. A GRAVITATING SPHERICAL SHELL Suppose, as in our earlier treatment of the problem, that we have a thin, uniform shell of matter, of radius R and mass M (Fig. 11-16). Let a particle of mass m be placed at some point P a distance r from the center of the shell. If we deal directly in terms of forces, then, as we saw, the force from material in the vicinity of a point such as A must be resolved along the line OP. In other words, a vector sum of all the force contributions is necessary. If we deal in terms of potential energy, however, we can cxploit its most important propcrty: Potential energy is a 446 Conservative forces and motion in space Fig. 11-16 Diagram for considering the gravitational potential energy due to a cir- cular zone ofa thin spherical shell of matter. scalar guantity. We can just add up the contributions to U from all parts of the shell. The force on m is then obtained simply from the relation F --™ tr dr (11-28) As in our direct calculation of the force, we take advantage of the symmetry of the system by considering the zone of material marked off on the sphere between the angles B and B + dB (Fig. 11-16). For this zone we have area = 2ir R sin B dd 2wR 2 sm9d6 mass 4jt/?2 M i.e., dM = iMsinddd All of this material in dM is at the same distance, s, from P. Thus the contribution that dM makes to U is given, according toEq. (11-27), by dU = - GmdM GMm sin 6 dB s 2 s The total potential energy of m is thus ... . GMm f sin 6 dB U(r) = - (11-29) 2 J s This integral is to be evaluated, keeping R and r constant, by 447 A «ravilating spherical shell allowing to sweep from zero to ir (s changing accordingly) so as to include the contributions of all parts of the shell. The calculation is made simple by the fact that, in the triangle AOP, we have s 2 = R 2 + r 2 - 2Rrcos0 Differentiating both sides with respect to 0, remembering that R and r are constants for the purpose of the integration in Eq. (11-29), wehave 2*^- = 2Rrsin6 do Thereforc, sin 6 dd ds^ Rr But the left-hand side of this is just the integrand of Eq. (1 1-29), which thus becomes U(r) = - GMm 2Rr 7»-o ds (11-30) Equalion (11-30) is the key result in cakulating gravitational potenlial energies andforces due lo spherkal objecls. In evaluating Fig. 11-17 (o) Gravi- tational potential energy of a point parlicle as a function of its distancc from the center of a thin splierical shell of radius R. (b) Varia- tion of F with r, de- riaed from (a). 448 Conscrvative forces and motion in space the integral of ds, however, we must distinguish two cases: Case 1. Point P outside the shell (i.e., r > R, as in Fig. 11-16). The limits on s are defined as follows: = giving s,„i n = r — R = 7r giving s mnx = r + R Hence Smnx Anln = 2*1 U(j.)--2MH (point o«to<fe shell) (11-31) Ca.se 2. Point P inside the shell (i.e., r < R). The limits now become 8 = giving s mi „ = /? - r 9 = x giving s msx = /? + r Hence U(r) = - ^^ (point inside shell) (11-32) These two results are disarmingly alike. The forces derived from them are, however, quite different. If we proceed now to calculate the forces from Eq. (11-28), we must remember that R (the radius of the shell) is a constant. It is only r, the distance of the mass m from O, that can vary. Hence we have Case 1. F r = j— (point outside shell) (11-33) Case 2. F r = (point inside shell) (11-34) In Fig. 11-17 we show the graphs of potential energy and forcc as functions of the radial distance r of the test mass m from the center of the shell. The discontinuities at r = R are easily seen. Once we have obtained these results, it is a simple matter to consider a solid sphere of material. 449 A gravitaling spherical shell A GRAVITATING SPHERE Our primary assumption is that the sphere can be regarded as made up of a whole succession of uniform spherical shells, even though the density may vary with radial distance from the center. Granted this assumption of symmetry, we can at once draw some conclusions about the gravitational force cxerted by such a sphere, of total mass M and radius R. Case 1. Observation point oulside the sphere (r > R). Since each component spherical shell acts as though its whole mass were at the center, the same can be said for the sphere as a whole. Regardless of the way in which the density of the ma- terial varies with radius, we have F(r) = — (point oulside sphere) (11-35) This is the result that we have already used in Chapter 8 in discussing Newton's famous comparison of the gravitational accelerations of the moon and the apple. Case 2. Observation point inside sphere (r < R). Here we must be more careful, but we can at once make two clear state- ments: a. For all the spherical shells lying oulside the observation point, the contribution to the force is zero, by Eq. (1 1-34). b. For all shells lying inside the radius defined by the ob- servation point, the mass is effectively concentrated at the center, byEq. (11-33). To specify the force in this case, therefore, we must know how much mass is enclosed within the sphere of radius r drawn through the observation point. Case l', (special) The same as case 2, but with the extra proviso that the density of the sphere is the same for all r — the sphere is homogeneous. In this case, we know that the fraction of the mass contained within radius r is equal to r 3 /R 3 , because this is the ratio of the partial volume to the whole volume. Hence the amount of mass to be considered is equal to Mr 3 /R 3 , effec- tively at the center, i.e., a distance r from the position of the test mass. This then gives us F ^ = W r (P°' nt inside homogeneous sphere) (11-36) 450 Conservative forces and mol ion in space Fig. 11-18 (a) Force on a poinl particle as a function of ils dis- lancefrom the center ofa uniform solid sphere of radius R. {b) Variation of mutual polential energy with r, obtained front {a) by integra- tion. The approxi- mately linear increase ofU with r jusi ouiside r = R corresponds to gravily as observed near the earih's sur- face. The combined results of Eqs. (11-35) and (11-36) for a homo- geneous sphere are shown in Fig. ll-18(a). One can also con- struct the graph of potential energy versus r for all values of the distance of a particle of mass m from the center of the sphere. This is shown in Fig. ll-18(b). For interior points (r < R) we have »-w"j," GMm 2R3 (R2 - r2) But we already know, by putting r = R in Eq. (11-31), that GMm U(R) = - Thus we get R (r<R) </(,)= -^(3K2-,2) In particular, at r = 0, we have (11-37) 451 A gravitating sphere 1/(0) = - 3GMm 2R If one imagines starting out from r = 0, then, as Fig. ll-18(b) shows, the potential energy increases parabolically with distance up to r = R and then goes over smoothly into the continued increase of U with r that is described by Eq. (11-31). The variations of F(f) and U(r) for r > R in Fig. 1 1-18 hold good, as wc have seen, for any spherically symmetrical distribu- tion of matter; there is no requirement that the sphere should be homogeneous. On the other hand, one must be careful to remember that Eqs. (11-36) and (11-37), for r < R, refer only to the special case of a sphere of the same density throughout. Thus it does not correctly describe the variation of gravitational force with radial distance inside such objects as the earth and the sun, which have drastic variations of density with r (see Fig. 11-19). This invalidates a favorite textbook cxercise: "Show that a particle would execute simple harmonic motion in a tunnel bored along a diameter of the earth." The practical impossibility of making such a tunnel may, however, be considered as a far more powerful objection. If one is used to thinking that the potential energy of an object of mass m, a distance h above the earth's surface, is given Fig. 11-19 (a) Radial variation of density inside the earth. (After E. C. Bullard.) (b) Calculated variation of density wiih radial distance inside the sun (I. Iben and Z. Abraham, M.I.T.). 452 Conservative forces and motion i n spacc by the formula U = mgh, it may seem hard to reconcile this result with the result expressed by Eq. (11-31): U(r)=-™^ (r>R) There is really no difficulty, however, once one recognizes that the zero of potential energy is arbitrary and that the simple linear formula applies only to objects raised through distances that are exceedingly small compared to the earth's radius. We have, in fact, from the more general formula [i.e., Eq. (11-31)], U(R + h)-U(R)=-^- + ^- R + h R GMm R2 ■h Since, however, G Mm/ R 2 is the gravitational force mg exerted on m, this gives us U(R + h) - U(R) « mgh By putting U(R) = as an arbitrary reference level of potential energy, we see that U = mgh is an acceptable approximation that applies to small displacements near the earth's surface. This approximatcly linear increase of potential energy with distance just outside a sphere is indicated on Fig. ll-18(b). ESCAPE VELOCITIES Suppose we have an object at the surface of a large gravitating sphere, such as a planet, and we want to shoot the object off into outer space so that it never returns. What speed must we give it? This problem is easily considered in terms of energy conservation. At the surface of the planet (r = R) the potential energy is given by U(R) = _ ^ R At r = co we have U = 0. The particle, to reach /■=<», must survive to this distance with some kinetic energy .K(oo). Thus at launch it must have a kinetic energy K(R) defined through the conservation equation 453 Escape velocities Therefore, GMm K(R)> R The critical condition is reached if we take the equality in the above statcmcnt; the object would then just reach infinity with zero residual speed. The minimum escape speed at radius R is thus given by , , GMm Therefore, * = ^ (11-38) In calculating the escape velocity from the earth's surface, it may be convenient to state the result of Eq. (11-38) in another form that makes use of our knowledge that the gravitational force on a mass m at r = Re can be set equal to the magnitude of w times the local acceleration due to gravity, g: GMm ~R^ =mg Therefore, GM and so, from Eq. (11-38), we have DO = VgR E ) v2 (H-39) Putting in the familiar values g ~ 9.8 m/sec 2 , R E ~ 6.4 X 10 G m, we have bo « 11 .2 km/sec Notice once again the remarkable implications of energy con- servation. For the purpose of calculating complete escape, wc need spccify nothing about the direetion in which the eseaping object is fired. It could be radially outward, or tangentially, or anything in between ; the same value of v u applics to all cases [Fig. ll-20(a)]. It is interesting that the magnitude of the escape speed v a is exactly \/2 times as great as the orbital speed that a partiele 454 Conservative forees and motion in space Fig. 11-20 (a) The possibility ofescape from a gravitating sphere depends on the magnitude of the velocity, not on its direction. (b) The escape speedfrom a double-star system depends on the position o f the start ing point P. would have if it could skim around the earth's surface in a circular orbit of radius Rj,;. The preceding calculation assumes that the object from which the escape takes place is isolated from all other objects. More complicated escape problems would arise if, for example, we wanted to calculatc the escape velocity from the surface of one member of a double-star system [Fig. 1 l-20(b)]. We should then have to consider where the Iaunching point P was in relation to the centers O t and 2 of the two stars. But still the scalar property of potential energy makes the calculation relatively simple. The potential energy of a mass m at P is just the straight- forward sum of the potential energies due to the two stars sep- arately. And again the direction of launeh for complete escape at the critical velocity is immaterial (as long as the trajectory misses the other star!) A particularly important example of this last type of system is, of course, the earth-moon combination. This is shown schematically in Fig. 11-21 (a). If we consider a partiele of mass m, its potential energy as a funetion of position along a straight line joining the centers of earth and moon is as shown by the full line in Fig. ll-21(b). Mathematically it is given by the formula U(r) = - GMf.iti GM.\ r m r ' D- r (11-40) where D is the distance between the centers of earth and moon and r is the distance of the partiele from the center of the earth. This potential energy is simply the sum of the separate con- tributions from earth and moon. 455 Escape velocilies Effi ■;[.-''■■- 'v BBS Fig. 11-21 (a) Sche- malic diagram of the earth-moon system. (b) Form ofvariation Ofthe total potential energy of a particle along Ihe line joining the centers of earth and moon. Moon The maximum value of U(f) occurs at the point r = r„, where (dU/dr) rm = 0. This is clearly also the point at which the attractive forccs due to earth and moon are equal and op- posite, so that we can put GMp.m CMsttn (.D - r m y This gives 1 + Wm/Mk) 11 * Now the value of M M /M E is about S V. so that r m w 0.90D. This represents a point about 24,000 milcs from the moon's center. The value of U(r) at this value of r is given, according to Eq. (11-40), by U(r,„) = - GM K m GM M m 0.923 G Me>» 0.9D 1. 23GM E m D 0AD + 3 1 ) If a spacecraft is to reach the moon from the earth along the 456 Conservative forccs and motion in spacc direct line considered, it would have to surmount this potential energy hump. If, further, this were to be done by simply coasting after an initial rocket blast at the point of departure from the earth's surface, the necessary minimum initial speed at the earth would be defined by the following equation: , GM E m ... l.23GM B m fwoo 2 ^— = U(r m ) = Uh D This then gives us 2GMk ( l.23R K \ The factor 2GMe/Re «s just the square of the speed needed for complete escape from the earth, as given by Eq. (11-39) and evaluated numerically as I1.2km/sec. Putting R E /D ~ g\y in the above expression for v 2 , one finds u « 11.1 km/sec. At this minimum velocity the traveling object would just barely surmount the potential "hill" between earth and moon and would then fail toward the moon, reaching its surface with a speed of about 2.3 km/sec. (You should check thesc results for yourself.) MORE ABOUT THE CRITERIA FOR CONSERVATIVE FORCES 1 In our discussions of forces and potential energy, we have pointed out that the fundamental criterion for a force to be conservative is that the net work done by the force be zero over any closed path. As an aside we pointed out that in one-dimen- sional problems, but only in one-dimensional problems, this condition is automatically met if the force is a unique function of position. We shall now give a simple example to illustrate how, in two dimensions, this latter condition is not sufficient. A force F(x, y) that is a unique function of x and y may never- theless be nonconservative. Our example is this: Suppose that a particle, at any point in the xy plane, finds itself exposed to a force F given by F x = -ky F„ = +kx where k is a constant and x and y are the coordinates of the 'This section may be omitted without loss of continuity. 457 More about ihc critcria for conservative forces Fig. 11-22 (a) Rectangular path in a plane. (b) Smooth tube shaped to force a parlicle tofollow the palit shown in (a), (c) Closed path made up oftwo different paths between llte giuen points A and B. If the force is conservalive, the net work is zero. particle. Such a force evidently dcpends only on the position of the particle. Now let us calculate the work done by this force on the particle if the latter moves counterclockwise around the closed path shown in Fig. ll-22(a). It may well be objected. that the particle would never follow this path under the action of the force F and nothing else, but (as we did in our first approach to potential energy at the beginning of Chapter 10) we can imagine that F is exactly balanced by another force, supplied for example by a spring, so that the object can be freely moved around without any net work being donc. Or, alternatively, we can imagine a very smooth pipe, as in Fig. 1 l-22(b), which com- pels the particle to travel along the sides of a rectangle and yet does no work on the particle because the force exerted by the pipeis always at right angles to the particle's motion. We should also add the proviso, in this case, that the particle has enough kinetic energy to carry it around the path even ifFistaking energy away from it. Now for the caleulation. Starting at O, let the particle move along the x axis a distance a to the point P with coordinates x = a, y = 0. The work done by F during this motion is W\ = / F x dx = -ky j dx = since y = everywhere on this portion of the closed path. Next, the particle moves from P to Q{x = a, y = b). The work done by F along this path is 458 Conservative forees and motion in space = / F y dy = +ka dy = kab Jv-0 Jo w» ly-0 For the path from Q to R (0, b), the work done is W 3 = / F x dx = -kb / dx = +kab and finally, the work W 4 done as the particle moves from R back to the origin O is zero, since x and hence F v is zero every- where on this path : W 4 = The total work done in the round trip is therefore W = W\ + W% + Ws + W a = Ikab * and the force F given by Eq. (1 1-41) is not conservative, although it depends only on position. If k were positive, the kinetic energy of the particle would be increased by 2kab for each complete circuit taken in this direction. On the other hand, if the particle traveled clockwise, its kinetic energy would be decreased by this amount each time. This may seem — and indeed is — a very artificial example; nevertheless, forces having precisely this non- conservative property [although not the same analytic forms as described by Eq. (1 1-40)] play an important role in physics, especially in electromagnetism. There is another way of looking at the analysis of such a situation. The physical criterion for the force to be conservative is that it should do no net work, either positive or negative, as the particle to which it is applied makes a complete circuit around the path of Fig. ll-22(a) in either direction. Let us consider a more general situation [Fig. ll-22(c)] in which a particle travels from a point A to a point B along one path and returns from B to A along a dirferent path (again imagine that constraining forces, doing no work, are applied as necessary). We shall assume that the force we are studying (i.e., not including any constraining forces) is a function of the position of the par- ticle only. If this force is conservative, we then have W = j F-ds+ F-ds = Path l P«th 2 459 More about the critcria for conservative forces It follows from this that / F • ds = - / F ■ cis Path 1 Path 2 But now, if the force is a function only of position, we can inter- change the limits of the integral on the right and reverse its sign. Hence, if F is conservative, we must have •B r B Fds 'A Path 1 Path 2 Path 1 Pi If this condition is satisfied we can put Wab=I Fds (ll-42a) without reference to the particular path. And if W AB is the same for all paths from A to B, we can conclude that the force is conservative and that the potential-energy function can be defined through the equation V b -Ua--J a F ds (ll-42b) In evaluating work integrals such as that in Eq. (ll-42a), one may wish to resolve the force F into its components F x and F„, and resolve the element of path ds into its components dx and dy, thus getting (for a two-dimensional problem) W \ F,dx+ F„dy If one uses this equation, however, one should always remember that, basically, the integrals are defined as being taken along an actual designated path of the particle. This might seem to be obvious, because the force must necessarily be applied wherever the particle is. However, there are many situations in which an object experiences a force wherever it happens to be (gravitational force, for example). One can then set up a purely mat hemat ical statement that defines a value of F for any given x and y. In general each component of F is a function of both x and y. And unless one already knows that the force is conservative, it is 460 Conservative forees and motion in ^pace FIELDS 1 Fig. 11-23 Consideration of work done dong a specified path between two points. essential to f'ollow the given path, as in Fig. 11-23, and not take the simpler but perhaps unjustified course of (for example) finding the integral of F x from jci to x 2 along the line AC (y — y\ = const.), followed by the integral of F y from y>\ to y 2 along the line CB (x = x 2 = const.). It is perhaps worth ending this discussion with the remark that, in many situations, one may usc the concept of potential energy even when additional nonconscrvative forces act on the particle. For example, a satellite moving in the gravitational field of the earth may be subject to the frictional drag of the earth's upper atmosphere. This drag is a dissipative force, non- conservative, and the total mechanical energy of the motion will not be constant but will decrease as the motion proceeds. Nevertheless, one may still properly talk of the gravitational potential energy that the satellite possesses at any given point. The forces that we have labeled gravitational, electric, and magnetic are "action-at-a-distance" forces. No apparent physical contact between interacting objects comes into play. For this kind of interaction, the idea of a field of force is most useful. To introduce this concept, consider the gravitational attraction of the earth for a particle outside it. The puli of the earth depends both on the mass of the attracted particle and on its location relative to the centcr of the earth. This attractive force divided by the mass of the particle being pulled depends only on the earth and the location of the attracted object. We can therefore assign to each point of space a vector, of magnitude equal to ■This section may bc omitted without loss of continuity. 461 Fielcis the earth's puli on a particle divided by its mass, and of direction identical with that of the attractive force. Thus we imagine a collection of vectors throughout space, in general different in magnitude and direction at each point of space, which define the gravitational attraction of the earth for a test particle located at any arbitrary position. The totality of such vectors is called a field, and the vectors themselves are called the field strengths or intensities of the field. In this example, the gravitational field strength g at a point Pis F The magnitude of this field is measured in units of acceleration (e.g., m/sec 2 ) and is given explicitly by the law of gravitation as g ._ C £E»_ 6 £e r (11 _43) where M is the mass of the earth, r the position vector of the point P relative to the center of the earth, and e r the unit vector in the direction of r. Equation (11-43) is valid at all points out- side the surface of an ideal spherical earth. One can generalize this for the field produced by an arbitrary distribution of matter, for which the field strength g as a func- tion of position describes quantitatively the gravitational field. The gravitational force exerted on an object of mass m by this field is then given by F = mg. This field description of forces is especially useful in specify- ing electromagnetic forces. The electric field produced by a charged particle or by a collection of such charged particles is described by the electric field strength or intensity S, where Z = * F is the vector force acting on a positive test charge of magnitude q, and £ depends on position. For electric fields produced by charges at rest, the situation is similar to the gravitational case. As an aid to visualizing the character of a force field, use is frequently made of the concept of a field line. Starting from an arbitrary point, we draw an infinitesimal line element in the direction of the field at that point. Being thus brought to a neighboring point in the field, we draw another line element in 462 Conservative forces and motion in space the direction of the field at the new point, and so on. In the limit of making the line elements vanishingly short, we obtain a smooth curve, the tangent to which, at any point, is the direc- tion of the field at that point. This construct— which is the field line— has no real existence, but a vivid picture of the properties of a force field can be obtained by drawing a whole collection of such lines. Hand-in-hand with the concept of a line of force goes that of a line or surface of constant potential energy. We shall there- fore discuss this briefly in the next section. EQUIPOTENTIAL SURFACES AND THE GRADIENT OF POTENTIAL ENERGY 1 Besides its utility in connection with the dynamics of conservative systems, the concept of potential energy enables us to describe conservative fields of force quantitatively in a relatively simple fashion. The reason is that we can use the simple scalar field of potential energy to calculate the relatively complicated vector field of force. This calculation proceeds as follows. Suppose we know the potential energy U of a test particle at each point of space; i.e., we have a relation of the form U =f(x,y,z), where the single-valued function f(x, y, z) depends on the par- ticular field of force under consideration. If we wish to know at what points of space the test particle will have a given value of potential energy, say £/ , we set U = U and obtain an equation of the form f(x, y, z) = const. = Uo This is the equation of a surface, and this surface is called an energy eguipotenlial surface. There exists a whole family of these equipotential surfaces, one for each value of U(,. Since, by definition, it requires no work to move our test particle from one point to another on the same equipotential surface, it follows that the lines of force are everywhere perpendicular to the equi- potentials. We should draw attention to a distinction between two quantities here. Just as we define field as the force per unit mass (or charge, etc.) so we define potential (<p) as the potential energy per unit mass (or charge, etc). To take the specific case of gravi- ■This section may be omilled without loss of continuity. 463 Equipotential surfaces; poiential-encrgy gradicnts tation, we thus have the following paired quantities: gravitational force F (newtons) with gravitational potential energy U (joules), and gravitational field g (m/sec 2 ) with gravitational potential <p (mVsec 2 ). The complete array of field lincs and cquipotential surfaces (or, in two dimensions, equipotential lines) provides a graphic picture of a complete field pattern. In two dimensions we see two sets of curves which, however complicated their appearance may be, are everywhere orthogonal to one another. The gravita- tional field of a spherical object has a simple pattern, in which the field lines are radial lines and the equipotentials are a set of concentric spheres [Fig. ll-24(a)]. The field pattern due to two spheres close together is far less simple [see Fig. ll-24(b)]. (For one way of constructing such a diagram, see Problem 1 1-26.) In making drawings of the equipotentials, it is often convenient Fig. 1 1-24 (a) Equipotenlials and field lines due t o a sphere with a \/r 2 force law. (b) Eauipolenlials and field lines due lo a syslem oflwo nearby spheres, of masses 2 M and M. 464 Conservative forces and motion in space to draw them for equal successive increments of the potential (or potential energy); this makes the picture very informative, just like a contour map — which in effect it is (see Fig. 11-24). To obtain a complete specification of the force field, we must be able to get the magnitude of the force vector, as well as its direction, at every point of space. Consider a test particle at a point P and let it be displaced an amount ds to a neighboring point. Its potential energy will change by an amount dU = -F-ds F,ds This may be written in the form F* - - ~ (11-44) as In words, the component of the force F in any direction equals the negative rate of change of potential energy with position in that direction. The spatial derivative on the right- hand side of Eq. (1 1-44) is called a direclional derivative, because its value depends on the direction in which ds is chosen at the point P. (Strictly speaking, we should be using the notation of partial derivatives: F, = —dU/ds.) If we move from P to a neighboring point on the same equipotential surface as that on which P lies, then dU/ds is zero for this direction. If, however, we move to a neighboring point not on the same equipotential surface as that containing P, dU/ds will be different from zero. That particular direction for which dU/ds has its maximum value at a given point defines the direction of the line of force at that point, and the magnitude of this maximum value of dU/ds is the magnitude of the vector force at the point in question. This maximum value of dU/ds, together with its associated direction, is called the gradient of the potential energy; it is a vector directed at right angles to the equipotential surface. In symbols, we write F = -grad U (11-45) To help clarify the idea of the gradient, consider a con- servative field in two dimensions. Here the equipotentials are lines, rather than surfaces as they would be in three dimensions. In Fig. 1 1-25 we show two equipotentials, one for U = U and one for U = l/ -f AU. Starting at P on U , we can move to the equipotential U + M/ by any of an infinite number of dis- 465 Equipotential surfaces; potential-encrgy gradienls Fig. 11-25 Different paths between two neighboring equipo- tentials. _^\pyui* StAT^l^' 1 >u. IH^HHHHHH placements. However, for a given change of potential energy AU, one moves along a line of force to attain this change in the shortest possible displacement As. The rate of change of po- tential energy with position is maximum for this direction and this is the direction of the gradient of potential energy. This is indicated in Fig. 1 1-25, where three directions from P are shown. It is clear that As is shorter in length than either As! or As 2 and hence that AU/As is larger than AU/As x or AU/As 2 . MOTION IN CONSERVATIVE FIELDS We now turn our attention to the problem of the motion of particles in a conservative field of force. We have of course already discussed a number of problems involving motion under gravity, but here we shall try to indicate the value of the energy method as it applies to more complex situations. If the only force acting on a particle is that due to the field, the law of conservation of mechanical energy provides a first step in solving for the motion. This law, expressed as $mo 2 + U(x,y,z) = E (H-46) where v 2 = v x 2 + v u 2 + v, 2 , does not, however, contain the whole story. It is the result of combining three statements of Newton's law as applied to the three independent coordinate directions, and the synthesis of these three vector equations into one scalar equation involves the discarding of information that in principle is still available to us. Given that U is a conservative potential, we can always find its gradient and hence the vector force [Eq. (I l— 45)]. With this, plus the initial conditions (values of r and v at t = 0) everything that one needs for a solution of the problem is provided. Our interest here, however, lics in taking advantage of the Conservative forees and motion i n. spacc energy-conservation equation as far as possible, and supple- menting it with whatever other information may be necessary. This extra information may take the form of the explicit use of one or two of the Newton's Iaw statements, as we shall see in the example about to be discussed. Or it may, in suitable cir- cumstances, be contained in a conservation law for quantities besides energy — in particular, angular momentum. The ex- ploitation of this latter property is one of the principal concerns of Chapter 13. As a good example of the methods we shall consider the motion of a charged particle in a combined electric and magnetic field. Suppose we have a pair of parallel metal plates mounted inside an evacuated tube and connected to a battery as shown in Fig. 11-26, so that a uniform electric field 8 (= V/d) exists between the plates. The plates are placed between the poles of a magnet that produces a uniform magnetic field B in a direction perpendicular to the page. We shall assume that electrons start out with negligible energy and velocity from the lower plate. This could, for example, be arranged by giving the lower plate a photosensitive coating and shining light onto it, as in a com- mercial phototube. Consider one electron of charge q = — e emitted at O. It will be accelerated vertically by a constant force equal to eV/d directed along the positive y axis, and will be deflected into the path indicated in the figure. Since the magnetic deflecting force, always acting perpendicular to the particle's velocity, does no work and hence does not directly affect the particle's energy, the statement of energy conservation can be written simply as follows: Fig. 11-26 Motion of an electron in vacuum under the combined action of electric and magnetic fields. Adi Motion in conservative fields B into the page *U4- Fig. 11-27 Analysis ofthe magnelic force into compo- nenlS associated with the separate x and y components 0/7. where we have chosen V [= — (eV/d)y] equal to zero for y = O, i.e., at the bottom plate where the kinetic energy at y = O is negligible. As statcd at the beginning of this section, we need more information to determine the motion. Since the electric force is directed along the y axis, the only x component of force on the electron is the x component of the magnetic force. To evaluate this force, think of the velocity vector of the electron at some point of its path resolved into x and y components (Fig. 11-27). The component of magnetic force associated with v x in this situation is parallel to the y axis and does not concern us here, but the velocity component v u gives rise to a magnetic force component F x given by and Newton's law of motion for the x component of motion is accordingly ev v B=m^ (11-48) Equations (11-47) and (11-48) can be solved for the motion of the particle, as follows: Since v„ = dy/dt, Eq. (11-48) can be written as eB dy = m do x or, integrated, mv x = eBy (\\-49) since v x = O when y = 0. Now we use the value of v x from Eq. (11-49) in the energy equation (11-47) and get 468 Conservative forees and mol ion in space nwy2 + T®V eV y = (11-50) This is of the form of the energy equation in one dimension; the total energy is zero and the effective potential energy U'(y) is given by U'(y) = - e -jy + hm ®" This effective potential-energy curve is shown in Fig. 11-28. It is precisely of the form of a harmonic oscillator potential (centered on a point at a certain positive value of y) and its characteristic angular frequency co is given by eB O) = — m This implies a very interesting result. Equation (11-50) and Fig. 1 1-28 show that, as y increases from zero, the kinetic energy of the electron increases to some maximum value and then de- creases again. Mathematically, Eq. (11-50) defines a maximum value of y given by putting v u = 0: m V y»** - eB 2d If this is less than the separation d between the plates (a con- dition that can be obtained, as the above equation shows, by making B large enough), the y displacement of the electron will perform simple harmonic motion with the angular frequency w and with an amplitude A equal to £.y max . Taking t = at the instant the electron leaves the lower plate, we can put Fig. 11-28 Effective potential-energy curve for the y component of motion in the arrangement of Fig. 11-26. 469 Mol ion in conscrvativc fields Fig. 11-29 Cycloidal path ofelectron be- tween charged parallel plates in a magnetic field. y = /1(1 - cosco/) (ll-51a) What about the x component of the motion? If we look at Eq. (1 1-49), we see that v x is proportional to y. Since y is always positive, v z is always in the positive x direction. When one couples this with the harmonic oscillation along y, one sees that the path of the electron is a succession of rabbit-like hops, as shown in Fig. 11-29. Specifically, we have, from Eq. (11-49), dx eB — y d t m y oiy where w is the same angular frequency, eB/m, that we introduced earlier. Thus, substituting the explicit expression for y from Eq. (ll-51a), we have dx dt = a>/4(l — cos tor) Integrating this and putting x = O at f = O we then find that .v = A(ut - sinwr) (11-51 b) Equations (11-5 la) and (ll-51b), taken together, show that the path of the electron is in fact a cycloid— just as if it were on the rim of a wheel of radius A rolling along the x axis. THE EFFECT OF DISSIPATIVE FORCES We mentioned earlier how the conservative properties of a force field may be useful even in circumstances in which dissipative forces are also present. The slow decay of the orbits of artificial earth satellites provides a particularly interesting example of this. We all know that a satellite placed in orbit a few hundred miles above the earth's surface will eventually come down. The 470 Conservative forces and motion in spacc Fig. 11-30 (a) Earth satellite spiraling inward as it loses energy. (b) Small element of the path sliown in (a). descent is, however, spread over many thousands of revolutions. Thus, although the actual path is a continuous inward spiral, as shown in Fig. 1 l-30(a), the motion during a single revolution is very nearly a closcd orbit, which we will take to be circular for simpiicity. The statement of Newton's law at any stage is thus given, with extremely little error, by the usual equation for uniform circular motion: F = GMm _ mo r* r Thus the kinetic energy at any particular value of r is given by (11-52) „ . 2 G Mm K = %mu = 2r This exposes a very curious feature. As the orbital radius gets smaller, the kinetic energy increases — in other words, the satellite speeds up. This happens in spite of the fact that the motion of the satellite is continually opposed by a resistive force. If there were no such force the orbital radius and the speed would re- main constant. We know, however, that the satellite must have lost energy, and the amount of this loss is well defined. This becomes clear when we consider the potential energy: U= - GMm 471 The eflcct of dissipative forccs The total energy, E, when the orbit radius is r, is given by E-K+V--^ (11-53) As r decreases, E becomes more strongly negative. We may note the following relationships in any such circular orbit: E= -K=\U (11-54) But still one may wonder: Why does the satellite accelerate in the face of a resistive force? Figure 1 l-30(b) suggests the answer. The line AB represents a small segment, of length As, of the path of the satellite. The starting point A lies on a circle of radius r; the end point B lies on a circle of radius r + Ar (Ar being actually negative). We greatly exaggerate the magnitude of Ar, for the sake of clarity. We then see that as the satellite travels from A to B, it feels a component of the gravitational attraction along its path. If the resistive force is R(p), the total force acting on the satellite in the direction of As is given by _ GMm F = — — cos a — R(v) Thus, in the distance As, the gain of kinetic energy, equal to the work F As, is given by AK = F As = ^^Ascosa - R(v)As In the first term on the right, we can substitute As cos a = — Ar In the second term on the right, we can substitute As = v Al If we also put AK = mu Av, we arrivc at the following equation: GMm . „, . , »10 Ao = — Ar — R(u)u At However, by diffcrcntiation of both sides of Eq. (1 1-52) we have GMm . nwAv = - -2rT~ Ar Usingthisresult, wecan substitute for the value of —(GMm/r 2 )Ar 472 ( onscrvairvc forees and motion in space in the previous equation, and we get nwAv = 2mv&o — R(v)vAt Hence Ao = R(v) At m This is a most intriguing result. The positive sign is not an error. The rate of increase of speed of the satellite is directly propor- tional to the magnitude of the resistive force that opposes its motion! The seeming paradox is resolved when we recognize that the resistance, by changing the orbital path from a circle to an inward spiral, acts as an agent that allows the gravitational force to do positioe work on the satellite, the amount of which is numerically twice as great as the amount of negative work done by the resistive force itself. GAUSS'S LAW 1 The particular case of the inverse-square field of force due to a point mass or charge is so important that we will append here a brief discussion of Gauss's Iaw, which is a compact and powerful statement of the inverse-square law. Since we are primarily con- cerned with mechanies here, we shall diseuss the problem in terms of the gravitational rather than the electric field. We begin with a very elementary observation. As we have seen, the gravitational field g due to a point mass M is given by the equation .. GM This field acts at every point of the surface of a sphere of radius r centered on M. If we use the convention that the positive direc- tion is radially outward, then the radial component of g multi- plied by the total surface area of the sphere gives us a quantity <f> defined as follows: <t> = 47rr 2 £ r = -4irGM We shall call <t> the flux of the gravitational field g through the surface of the sphere. ■This seetion may be omitted without loss of continuity. 473 Gauss's law Fig. 11-31 (a) Gaussian surface enclosing a parlicle of mass M. There is a net gravitational ftux due to M. (b) Gaussian surface nol enclosing M'. The net flux due to M' is zero. Now a noteworthy property of <j> is that it is indcpcndent of r. If we had two spheres of different radii, r x and r 2 , centered on M, the flux of g would bc the same through both. We can extend this result to the case of a closed surface of any shape surrounding the mass [Fig. 11-31 (a)]. If at an arbitrary point on this surface the outward normal to the surface makes an angle B with r, as shown, we can rcsolve g into orthogonal components, parallel to the surface and along the normal. If we think of the flux literally as a kind of flow of the g field aeross the surface, it is only the normal component that contributes. Multiplying this by the element of area dS, we thus have d<$, = - — cos e dS Now dS cos 6 can be equated to the projection of dS perpen- dicular to r, and the quoticnt dS cos O/r 2 is the element of solid angle, dil, subtended at M by dS. Thus we can put d<f> = -GMcKl The total flux of g through the surface, as defined in this way, is then obtained by integrating over all the contributions dQ. But this means simply including the complete solid angle, Air. Thus we have -- I ,','o -GM dtt = -4irGM 474 Conservative forees and motion in spacc exactly as for a sphere. If we consider the situation for a point mass M' outside the surface, the result is quite different. This time a cone of small angle dti intersects the surface twice; the contribution to the flux is given by d<t> = -G M' 'dSi cos 6i dS 2 cos 2 n* r& It is not hard to verify that the two terms inside the parentheses are equal to — dQ and -\-dQ, respectively, so that d<t> = 0, from which it follows that the total flux <f> is also zero in this case. One can formalize these calculations a little by representing an element of surface area as a vector, dS, with a magnitude equal to dS and a direction along the outward normal, as shown in Fig. 11-31. The element of flux, d<f>, is then defined as equal to the scalar product g • dS. It is then apparent that, in Fig. 11-31 (b), gi • dSi and g 2 • dS 2 are of opposite sign. Suppose now that we have a number of masses, mu m 2< mz, . . ■ , inside a surface of some kind, as in Fig. 1 l-32(a). Their gravitational fields combine to produce a resultant gravita- tional field g at any point P: g = gi + g2 + g3 -\ The total gravitational flux is then given by <t> = J g • dS and we see that this is simply the sum of the contributions from the individual masses: Fig. 11-32 (a) Basis of calculating the gravitational flux due to an arbitrary collec- tion of masses inside a given surface. (b) The addition of external masses does not change the net flux, although it will alter the local field intensi- ties. 475 Gauss's law ■ / gi • dS + I g 2 • dS H- / g 3 • rfS + = —4irG(mi + ntz + t»i + • • •) Thus this total flux depends only on the total enclosed mass, regardless of how it may be distributed : <t> = — 4irG A/ to t a i (11-55) If, as in Fig. ll-32(b), we suppose that additional masses m[, m' 2 are placed outside the surface, our previous calculation shows that these contribute nothing to the total gravltational flux through the surface. Thus Eq. (11-55) emerges as a com- pletely general result; it is known as Gauss's law (or theorem), and its validity depends completely on the fact that the law of force is an exact inverse-square law. An exact parallel to it holds for electrostatics, in which context Gauss, in fact, first developed it. If a physical system has certain obvious symmetries, Gauss's theorem leads at once to important conclusions about field strengths. We shall consider a few examples, some of which we have already treated by other methods. APPLICATIONS OF GAUSS'S THEOREM Field outside a sphere Suppose we want to know the gravitational field at a point P, outside an isolated sphere of mass M at a distance r from its center [Fig. 1 1— 33(a)]. If the mass distribution within the sphere Fig. 1 1-33 Use of Gauss's law to calcu- lale the gracitalional field of a solid sphere (a) at exterior points, and (b) at interior points. 476 Conservative forccs and motion in space is symmetrical— i.c, if the density of matter is the same at all points at the same distance from the center of the sphere— then the gravitational field g itself has the same strength at all points on a spherical surface of radius r. Thus, if we draw a "Gaussian surface" in the form of a sphere of radius r, the total gravita- tional flux is given by # = 4irr 2 g r = -A-kGM whence GM g r = - -5- at once The result is of course entirely familiar, but the point to notice is that, once we have Gauss's general theorem, the process of carrying out an explicit integration over the mass distribution, as we did in Chapter 8, becomes unnecessary. Field inside a sphere An exactly similar treatment holds for the other familiar problem of the gravitational field inside a spherically symmetric mass distribution [Fig. 11— 33(b)]. Gauss's theorem tells us at once that, if we imagine a spherical surface of radius r < R, the material outside this surface contributes nothing to the flux of g through the surface. Thus, if the mass enclosed within the surface of radius r is m(r), we can at once deduce that Gm(r) The assumption of spherical symmetry throughout the system, outside as well as inside the radius r, is, however, essential. Otherwise the value of g may be different at different points on a spherical surface, and even though the total flux still has no contribution from exterior material, the field strength at in- dividual points will be affected. Thus we must assume, as in the first example, that the sphere is effectively isolated, i.e., far from any outside masses. Field due to afiat sheet Suppose we have an infinite sheet of matter, of surface density <r (mass per unit area). Symmetry in this case tells us that the field must be everywhere normal to the sheet and that it has the same 477 Applications of Gauss's theorem Fig. 11-34 Use ofGauss's law to calculate the gravi- talional field ofaftat sheet ofmatler. magnitude on the two sides (Fig. 11-34). We can construct a Gaussian surface for this problem in the form of a right cylinder, with ends of arbitrary shape but the same area A. Then there is no flux of g through the curved sides of the cylinder. The enclosed mass is just a A, so the total gravitational flux passing through the surface is — AtrGaA. But, by Gauss's theorem, this is equal to the field strength g multiplied by the total area of the two ends. Thus we have 2 Ag = -A-kGiA Therefore, g = — 2ttGV (11-56) The strength of the field is thus entirely independent of the distance from the sheet of matter, assuming this to be indeed of unlimited extent. In terms of gravitational systems, this example is not very realistic. But the result is highly relevant in electrostatics, where distributions of electric charge over large plane areas are fre- quently met. The parallel-plate capacitor is a prime example, and the constant field at all points between the charged plates in the Millikan experiment, for example, can at once be under- stood in the light of the above calculation. PROBLEMS 11-1 A particle of mass m slides without friction on a wire that is curved into a vertical circle (see the figure). The position of the particle can be described in lerms of 0, or in terms of the arc length s, or by the vertical distance h that it has fallen. (a) Consider the force that acts on the particle along the direc- tion of its path at any point, and show that the work done on the particle as it moves through an arc length ds is given by dW = F t ds = mgR sin d dd (b) By integrating this expression from d = to 6 = do, show that the work done is W = mgR(\ — cos 0o), and express this result in terms of the vertical distance h that the particle has descended. 11-2 Two blocks, of masses m and 2m, rest on two frictionless planes 478 Conservative forccs and motion in space inclined at angles of 60° and 30°, respectively, and are connected by a string through an eyelet of negligible friction (see the figure). Using energy methods, find the magnitude of the acceleration of the masses, and deduce the tension in the connecting string. 11-3 Show that if a mass on the end of a string is allowed to swing down in a circular arc from a position in which the string is initially horizontal (and the mass is at rest), then at the lowest point of the swing the tension in the string is three times as great as when the mass simply hangs there. (This result is quoted by Huygens at the end of his book on pendulum clocks, published in 1673.) 11-4 A pendulum bob of mass m, at the end of a string of length /, starts from rest at the position shown in the figure, with the string at 60° to the vertical. At the lowest point of the arc the bob strikes a previously stationary block, of mass nm, that is on a frictionless hori- zontal surface. The collision is perfectly elastic. (a) What is the specd of the bob just before the impact occurs? (b) What is the tension in the string at this instant? (c) What velocity is given to the block by the impact ? (d) In the oscillations of the pendulum after the collision, what maximum angle 6 to the vertical does the string make? (Obtain your answer in the form cos 6 = • • • .) 11-5 A student is assigned a "conservation of momentum" experi- ment in which two masses, mi and m% are suspended from strings of length /. Mass m\ is pulled aside a horizontal distance Xi and released. It strikes m? and sticks to it, after which the combined mass swings out to some final position x/. The student objects that mo- mentum is obviously not conserved in this proccss. It is zero initially, increases, and then decreases again to zero. The student also claims that energy is not conserved. Analyze carefully what actually goes on in this experiment, so that you are in a position to answer the student's objections. 11-6 A small ball of putty, of mass m, is attached to a string of length / fastened to an upright on a wooden board resting on a hori- zontal table (see the figure). The combined mass of the board and upright is M. The friction coefficient between the board and the table 479 Problems is fi. The ball is released from rest with the string in a horizontal position. It hits the upright in a completely inelastic collision. While the ball swings down, the board does not move. (a) How far does the board move after the collision? (b) What is the minimum value that n must have to prevent the board from moving to the lcft while the ball swings down? Assume that m <3C M, which gives the critical condition at 6 ~ 45°. 11-7 An object of mass m slides on a frictionless loop-the-loop apparatus. The object is released from rest at a height h above the top of the loop (see the figure). (a) What is the magnitude and direction of the force exerted on the object by the track as the object passes through the point A ? (b) Draw an isolation diagram showing the forces acting on the object at point B, and find the magnitude of the radial acceleration at that point. (c) Show that the object must start from h > r/2 to success- fully complete the loop. (d) For h < r/2 the object will begin to fail away from the track before reaching the top of the circle. Show that this happens at a position such that 3 cosa = 2 -|- 2h/r, where a is the angular dis- tance from the (upward) vertical. 11-8 A ball of mass m hangs at the end of a string of length /. It is struck so as to start out horizontally from this position with a speed vo- (a) What must Do be if the ball is just able to travel through a complete circular path ? (b) If l-o = \/4g/ (so that the ball could just rise through the height 2/ if projected vertically), what angle does the string make with the vertical when the ball begins to fail away from its circular path? (c) In situation (b), analyze the subsequent motion, assuming no losses of total mechanical energy. 11-9 A mass slides down a very smooth curved chute as shown. At what horizontal distance from the end of the chute does it hit the ground ? 480 Conservativc forces and motion in spacc 11-10 A daredevil astronomer stands at the top of his observatory dome (see figure) wearing roller skates, and starts with negligible velocity to coast down over thc dome surface. (a) Neglecting friction, at what angle does he leave the dome's surface? (b) If he were to start with an initial velocity t»o, at what angle would he leave the dome ? (c) For thc observatory shown, how far from thc base should his assistant position a net to break his fail, in situation (a)? Evaluate your answer for R = 8 m, and use g =* 10 m/sec 2 . 11-11 A pendulum of length L is held at an angle a from the vertical. A peg is located in the path of swing of the string as shown in the figure. The pendulum is released from rest. (a) Ifa = 60°,/3 = 30°, and r = Ly/3/4, what is the maximum angle with the vertical that the bob will reach ? (b) Show that the string will bucklc during thc bob's aseent when < (rcos/3 — Lcosa) < 3(L — r)/2. At what angle B does this buckling oecur? 481 Problcms m.q m,q (c) A carnival skill game involves such a pendulum and peg apparatus. A prize is awarded for causing the bob to strike the peg. In the game, L, r, and are fixed ; the contestant releases the bob from rest at any desired a. Find the value of a for which a prize will be won. (Express your answer in the form cosor = • • • .) 11-12 A frictionless airtrack is deformed into the parabolic shape y = \Cx 2 . An object oscillates along the track in almost perfect simple harmonic motion. (a) If the period is 30sec, by what distance is the track at x = ±2 m higher than at x = 0? (b) If the amplitude of oscillation is 2 m, how long does it take for the object to pass from x = —0.5 m to x = +1.5 m traveling in the positive x direction ? (The description of SHM as a projection of uniform circular motion is useful for this.) 11-13 A pendulum clock keeps perfect time at ground level. Ap- proximately how many seconds per week would it gain or lose at a height of 20 m above this level? (Assume a simple pendulum, with T = 2Tr\/I/g. Earth's radius = 6.4 X 10 6 m. 1 week ~ 6 X 10 5 sec.) 11-14 Two identical, and equally charged, ping-pong balls are hung from the same point by strings of length L (see the figure). (a) Given that the mass of each is m and that the equilibrium position is as shown in the figure, what is the charge on each? (Recall that the electrostatic force between two charges at a separation r is given by F = kqiq2/r' 2 .) (b) Suppose that the balls are displaced slightly toward each other by the same amount. Make appropriatc approximations and describe the subsequent motion. Find an expression for the frequency of oscillations. (c) On the other hand, if the ping-pong balls are displaced slightly in the same direction and by the same amount, what will the subsequent motion be? In this second case, does it matter whether they are charged or mcrely held apart by a massless rod? 11-15 (a) Show that free fail in the earth's gravitational field from infinity results in the same velocity at the surface of the earth that would be achieved by a free fail from a height H = R B (= radius of 482 Conservative forces and motion in space earth) under a constant acceleration equal to the value of g at the earth's surface. (b) Show that the speed at the earth's surface of an object dropped from height,/; (/i « R K ) is given approximately by | = v - (2*A)i« H i) (c) Verify the statement in the text (p. 454) that the speed needed for escape from the surface of a gravitating sphere is t'o\/2, where co is the speed of a particle skimming the surface of the sphere in circular orbit. 11-16 A physicist plans to determine the mass of the earth by ob- serving the changc in period of a pendulum as he deseends a mine shaft. He knows the radius R of the earth, and he measures the den- sity p s of the crustal material that he penetrates, the distance h he deseends, and the fractional change (AT/T) in the pendulum's period. (a) What is the mass of the earth in terms of these measure- ments and the earth's radius? (b) Suppose that the mean over-all density of the earth is twice the mean density of the portion above 3 km. (This supposition is actually rather accurate.) How many seconds per day would a pendulum clock at the bottom of a deep mine (3 km) gain, if it had kept time accurately at the surface? 11-17 The figure shovvs a system of two uniform, thin-walled, con- centric spherical shells. The smaller shell has radius R and mass M, the larger has radius 2/? and mass 2M, and the point P is their common center. A point mass m is situated at a distance r from P. (a) What is the gravitational foree F(r) these shells exert on m in each of the three ranges of r: < r < R, R < r < 2R, r > 2R? (b) What is the gravitational potential energy of the point mass m when it is at P? (Take the potential energv to be zero when m is infinitely far from P.) (c) If the particle is released from rest very far away from the spheres, what is its speed when it reaches P'! (Assume that the particle can pass freely through the walls of the shells.) 11-18 Assume the moon to be a sphere of uniform density with radius 1740 km and mass 7.3 X 10 22 kg. Imagine that a straight smooth tunnel is bored through the moon so as to connect any two 483 Problems points on its surface. (a) Show that the motion of objects along this tunnel under the action of gravity would be simple harmonic. (b) Calculate the period of oscillation. (c) Compare this with the period of a satellite traveling around the moon i n a circular orbit at the moon's surface. 11-19 The escape specd from the surface of a sphere of mass M and radius R is v e = (2GM/R) 1 ' 2 ; the mean speed v of gas molecules of mass m at temperature T is about (3kT/m) l/2 , where k (= 1.38 X 10 -23 J/°K) is Boltzmann's constant. Detailed calculation shows that a planetary atmosphere can retain for astronomical times (10 9 ycars) only those gases for which D J$ 0.2y e . Using the data below, find which, if any, of the gases H2, NH3, N2, and CO2 could be retained for such periods by the earth; by the moon; by Mars. M E = 6.0 X 10 24 kg, R E = 6.4 X 10 3 km, T B = 250°K. For the moon, M = Mu/Si, R = 0.27Rn, and T = T K . For Mars, M = O.llAfe, R = 0.53* s , and T - 0.8T E . 11-20 A double-star system consists of two stars, of masses M and 2M, separatcd by a distance D center to center. Draw a graph showing how the gravitational potential energy of a particle would vary with position along a straight line that passes through the centers of both stars. What can you infer about the possible motions of a particle along this line? 11-21 A double-star system is composed of two identical stars, each of mass M and radius R, the centers of which are separated by a distance of 5R. A particle leaves the surface of one star at the point nearest to the other star and escapes to an effectively infinite distance. (a) Ignoring the orbital motion of the two stars about one another, calculate the escape speed of the particle. (b) Assuming that the stars always present the same face to one another (like the moon to the earth), calculate the orbital speed of the point from which the particle is emitted. How would the exis- tence of this orbital motion affect the escape problem? 11-22 The earth and the moon are separated by a distance D = 3.84 X 10 8 m. Ignoring their motion about their common center of mass, but taking into account both of their gravitational fields, answer the following questions: (a) How much work must be done on a 100-kg payload for it to reach a height 1000 km above the earth's surface in the direction of the moon ? (b) Calculate the diiTerence in potential energy for this mass between the moon's surface and the earth's surface. (c) Calculate the necessary initial kinetic energy to get the payload to the moon. It must be greater than the potential difference 484 Conservalivc forces and motion in space given in (b) because the payload must overcome a "potential hill." Find the location of the top of this hill and compute the potential difference from the earth to it. 11-23 Two stars, each of mass M, orbit about their center of mass. The radius of their common orbit is r (their separation is 2r). A planetoid of mass m (« M) happens to move along the axis of the system (the line perpendicular to the orbital plane which intersects the center of mass), as shown in the figure. (a) Calculate directly the force exerted on the planetoid if it is displaced a distance z from the center of mass. (b) Calculate the gravitational potential energy as a function of this displacement z and use it to verify the result of part (a). (c) Find approximate expressions for the potential energy and the force in the cases zy> r and z « r. (d) Show that if the planetoid is displaced slightly from the center of mass, simple harmonic motion occurs. Compare the period T P of this oscillation with the orbital period 7"o of the binary system. 11-24 A frictionless wire is stretched between the origin and the point x = a, y = b in a horizontal plane. A bead on the wire starts out from the origin and moves under the action of a force that varies with position. Find the kinetic energy with which the bead arrives at the point (a, b) if (a) The components of the force are F x = k\x, F y = kiy- (b) The components of the force are F x = kiy, F„ = A: 2* (with ki, *2 >0). In one case the force is conservative, in the other case it is not. By considering a different path — e. g., from (0, 0) to (a, 0) and then from (a, 0) to (a, b) — verify which is which. 11-25 The nuclear pari of the interaction of two nucleons (protons or neutrons) is described pretty well by a potential V(r) = —\e~ r/r o/r when the separation r is greatcr than 1 F (1 F = 10~ ls m). In this expression, /-o = 1.4 F and X = 70 MeV-F. (The potential was proposed by H. Yukawa and is named after him.) (a) Find an expression for the nuclear force that acts on (each of) two nucleons separated by r > 1 F. (b) Evaluate this force for two protons 1.4 F apart, and com- pare this with the repulsive Coulomb force at that separation. (c) Estimate the separation at which the nuclear force has 485 Problem s dropped to 1% of its value a t r = 1.4 F. What is the Couiomb force at this separation? (d) The last result indicates why the Yukawa potential is not part of our macroscopic experience. Note that ro characterizes the range of distance over which the interaction is important. In contrast to the nuclear force, gravitational and Couiomb forces have been called forces of infinite range. Indicate why such a description is appropriate. 11-26 In Fig. ll-24(b) is shown a plot of gravitational equipotentia!s and field lines for two spheres of masses 2M and M. One way of developing this diagram is as follows. We know that the gravitational potential at any point P is given by G(2M) G(M) where n and r 2 are measurcd from the centers of the spheres. An equipotential is defined by <pp = constant. In order to construct such equipotcntials, draw a set of circles, with the sphere of mass M as center, such that the values of 1/ra are in arithmetic progression — e.g., IA2 = 1, 1.5, 2, 2.5, 3, 3.5 Then with the sphere of mass 2M as center, draw a set of circles with radii r\ twice as great as the values of r 2 . Since <p P ~ (2/n) + (IA2), the corresponding circles in each pair represent equal contributions to <p by 2M and M. The two sets of circles have a number of intersections, and each intersection corresponds to a well-defined value of <fp. For instance, with 10r2 = 2.85 units and 10n = 10 units, we have <pe ~ (2/1.0) + (1/0.285) = 5.5. But <pp ~ 5.5 is obtained also at the intersections of other pairs of circles as follows: r 2 = 0.33, n = 0.80; r 2 = 0.40, n = 0.67; /•2 = 0.67, n = 0.50. Joining these intersections having the same value of (2/n) + (1/ra) gives us an equipotential, and other equi- potentials can be constructed in the same way. Once the equipo- tentials are obtained, the field lines can be drawn as lines everywhere normal to the equipotentials. 11-27 A science fiction story concerned a space probe that was re- trieved after passing near a neutron star. The unfortunate monkey that rode in the probe was found to be dismembered. The conclusion reached was that the strong gravity gradient (dg/dr) which the probe had experienced had pulled the monkey apart. Given that the probe had passed within 200 km of the surface of a neutron star of mass 10 33 g (about half the sun's mass) and radius 10 km, was this explana- tion reasonable? (The probe was in free fail at the time.) 11-28 A straight chute is to be used to transport articles a given horizontal distance /. The vertical drop of the chute can be freely chosen. The articles are to arrivc at the top of the chute with negligiblc 486 Conservativc forces and motion in. space velocity and the chute is to be chosen such that the transit time is a minimum. (a) If the surfacc is frictionless, what is the angle of the chute for which the time is minimized? (b) What is the corresponding angle if the coefficient of fric- tion is m? (c) If a playground slide were designed to give minimum dura- tion of ride for given horizontal displacement, and if the coefficient of friction of the child-slide surface is 0.2, what angle would the slide make with the horizontal? (Ignore the curved portion at the lower end of the slide.) (d) If the optimizaiion problem of (a) is encountered and if curved chutcs are allowed, can you guess roughly what form the best design would have? 11-29 The magnetic force exerted on a particle of charge q and mass m traveling in a uniform magnetic field B (which is constant in time) is given by F = kqs X B (where the constant k = 1 in the MKS system). (a) Show that the work done by such a force on the particle is zero for arbitrary particle motion. (b) Situations arise in which a moving charged particle is slowed by a force of constant magnitude F D and of direction opposite to the instantaneous velocity. (For example, the ionization of atoms along the track of a charged particle produces an almost constant energy loss per unit distance, corresponding to a constant retarding force.) Show that if a particle has speed eo at / = 0, thcn for / > its speed will be given by u(t) = vo — Fnt/tn until t = /huo/Fd, after which time the particle will remain motionless. Note that the presence of a magnetic field will not affect this result. (c) Under the action of a magnetic force alone, a particle with an initial velocity vo (normal to B) describes a circle of radius r = moo/kgB with a circular frequency o = kgB/m. Show that a particle subject also to the force described in (b) moves inward along a spiral and that the number of circuits it makes in the spiral before it comes to rest is given by {kqBoa/2irFti). 11-30 (a) Obtain an expression for the gravitational field due to a thin disk (thickncss d, radius R, and density p) at a distance h above the center of mass of the disk. In calculating the field, assume the surface density a = pel to be concentrated in a disk of negligible thickness a distance h beneath the test point. This will give an accurate result whenever h^> d, and reduces the complexity of the caleulation. Note that for R — * « , the caleulation agrees with the predietion of Gauss's theorem, as it must. (b) Express the field obtained in part (a) as a fraetion of 2irGcr for the cases R = 2h, R = 5h, and R = 25h. (c) How many seconds per year would a pendulum clock gain 487 Problcms when suspended with the bob 5 cm above a lead floor 1 cm thick, if it keeps correct time in the absence of the floor? 11-31 Any mass of mattcr has a gravitational "self-energy" arising from the gravitational attraction among its parts. (a) Show that the gravitational self-energy of a uniform sphere of mass M and radius R is equal to -3GM 2 /5R. You can do this either by calculating the potential energy of the sphere directly, inte- grating over the interactions between all possible pairs of thin spherical shells within the sphere (and remembering that each shell is counted twice in this calculation) or, perhaps more simply, by imagining the sphere to be built up from scratch by the addition of successive layers of matter brought in from infinity. (b) Calculate the order of magnitude of the gravitational self- energy of the earth. Check where this lies on the scale of energies displayed in Table 10-1. 488 Conservativc forces and motion in space Part III Some special topics Whenfirst studying mechanics one has the impression that everything in this branch ofscience is simple, fundamental and sett/edfor all time. One would hardly suspect the existence ofan important clue which no one noticedfor three hundred years. The neglected clue is connected with one of the fundamental concepts of mechanics — that ofmass. A. EINSTEIN AND L. INFELD, The Evolution ofPhysics (1938) 12 Inertial forces and noninertial frames imagine that you are sitting in a car on a very smooth road. You are holding a heavy package. The car is moving, but you cannot see the speedometer from where you sit. Ali at once you get the feeling that the package, instead of being just a dead weight on your knees, has begun to push backward horizontally on you as well. Even though the package is not in contact with anything except yourself, the effect is as if a force were being applied to it and transmitted to you as you hold it still with respect to yourself and the car. If you did not restrain the package in this way, it would in fact be pushed backward. You notice that this is what happens to a mascot that has been hanging at the end of a previously vertical string attached to the roof of the car. How do you intcrpret these observations? If you have any previous experience of such phenomena, you will have no hesita- tion in saying that they are associated with an increase of velocity of the car — i.e., with a positive acceleration. Even if this were your first experience of this type, but if you had a well-developed acquaintance with Newton's laws, you could reach the same conclusion. An acceleration of the car calls for an acceleration of everything connected with it; the acceleration of the package requires, through F = ma, a force of the appropriate size supplied by your hands. Nonetheless, it does feel just as if the package itself is somehow subjected to an extra force— a "force of in- 493 ertia" — that comes into play whenever the effort is made to change the state of motion of an object. These extra forces form an important class. They can be held responsible for such phenomena as the motion of a Foucault pendulum, the effects in a high-speed centrifuge, the so-called g forces on an astronaut during launching, and the preferred direction of rotation of cyclones in the northern and southern hemispheres. These forces are unique, however, in the sense that one cannot trace their origins to some other physical system, as was possible for all the forces previously considered. Gravita- tional, electromagnctic, and contact forces, for example, have their origins in other masses, other charges, or the "contact" of another object. But the additional forces that make their ap- pearance when an object is being accelerated have no such physical objects as sources. Are these inertial forces real or not? That question, and the answer to it, is bound up with the choice of reference frame with respect to which we are analyzing the motion. Let us, therefore, begin this analysis with a feminder of dynamics from the standpoint of an unaccelerated frame. MOTION OBSERVED FROM UNACCELERATED FRAMES An unaccelerated reference frame belongs to the class of reference frames that we have called inertial. We saw, in developing the basic ideas of dynamics in Chapter 6, that a unique importance and interest attaches to these frames, in which Galileo's law of inertia holds. We have seen how, if one such frame has been identified, any other frame having an arbitrary constant velocity relative to the first is also inertial, and our inferences about the forces aeting on an object are the same in both. To a good first approximation, as we know, the surface of the earth defines an inertial frame. So also, therefore, does any system moving at constant speed over the earth. Galileo himself was the first person to present a elear recognition of this faet, and one aspect of it that he diseussed is useful as a starting point for us now. In his Dialogue Concerning the Two World Systems, in which he advocated the Copernican view of the solar system in preference to the Ptolemaic, Galileo pointed out that a rock, dropped from the top of the mast of a ship, always lands just at the foot of the mast, whether or not the ship is moving. Galileo argued from this that the vertical path of a falling object does not compel one to the conclusion that the earth is stationary. 494 Inertial forces and nonineriial frames Fig. 12-1 (a) Para- bolic trajectory under grauity, as observed in the eanh's reference frame. The initial velocily v o is hori- zontal, (b) Same motion observed from a frame wilh a hori- zontal velocity greater than vo. (c) Same motion observed from a frame having bot h horizontal and vertical velocily components. The comparison here is between an object falling from rest relative to the earth and another object falling from rest relative to the ship. If we considered only an object that starts from rest rela- tive to a moving shtp, its path would be vertical in the ship's frame and parabolic in the earth's frame. More generally, if we considered an object projected with some arbitrary velocity relative to the earth, its subsequent path would have diverse shapes as viewed from different inertial frames (see Fig. 12-1) but all of them would be parabolic, and all of them, when ana- lyzed, would show that the falling object had the vertical ac- celeration, g, resulting from the one force F„ (= mg) due to gravity. Let us now contrast this with what one finds if the reference frame itself has an acceleration. MOTION OBSERVED FROM AN ACCELERATED FRAME Suppose that an object is released from rest in a reference frame that has a constant horizontal acceleration with respect to the earth's surface. Let us consider the subsequent motion as it appears with respect to the earth and with respect to the ac- celerating frame. We shall take the direction of the positive x axis in the direction of the acceleration and will set up two rectangular coordinate systems: system S, at rest relative to the earth, and S', fixed i n the accelerating frame (Fig. 12-2). Take + v Fig. 12-2 Relationship of co- ordinales ofa particle in iwo frames that are in accel- erated relative motion. T s- (relative to 5) —fP O ■+x 495 Motion observed from an accelerated frame the origins of the frames to coincide at t = 0, and suppose that the velocity of S' with respect to S at this instant is equal to Vo- The vertical axes of the two systems are taken as positive upward, and the object is released at t = from a point for which x = x' = 0, y = / = h. What will the trajectories in S and S' look like? For an observer in S, we already know the answer. To him, the object is undergoing free fail with initial horizontal velocity v [Fig. 12-3(a)]. Thus we have (As observed i n S) {x = vot y- h- i** 2 These two equations uniquely define the position of the object at time t, but to describe the motion as observed in S' we must express the results i n terms of the coordinates x' and y' as mea- sured in S'. To transform to the S' frame, we substitute x = x — x, y = y where x s is the separation along the x axis of the origins of S and S' (see Fig. 12-2). We know that x, = vot + §at 2 Substituting these values we find .au a- ca K = Vot " ( " ' + % a,2) = ~% a ' 2 (As observed in S') \ , „ W = h - \gi- Thus the path of the particlc as observed in S' is a straight line given by the equation x'= -"{h- y') 8 Fig. 12-3 (a) Para- bolic trajectory of a particle under gravily, as observed in the earlh's reference frame S. (b) Same motion observed in a frame S' that has a constant horizontal acceleration. . . . . : O ^s N \ - \ Seen in S (a) 496 Inertial forces and noninertial frames This is shown in Fig. 12-3(b). In the accelerated frame, the object appears to have not only a constant downward component of acceleration due to gravity, but also a constant horizontal component of acceleration in the — x direction which causes the particle to follow a nonvertical straight-line path. [A similar simple example is the monkey-shooting drama described in Chapter 3 (p. 104). The outcome becomes almost self-evident if we choose to describe the events in the rest frame of the falling monkey. In this frame the bullet just follows a straight-line path directly toward the monkey, while the ground accelerates up- ward at 9.8 m/sec 2 .] There is no mystery about the unfamiliar motion repre- sented by Fig. 12-3(b). It is a direct kinematic consequence of describing the normal free-fall motion from a frame that is itself accelerated. We could perfectly well use this path, de- scribed by measurements made entirely within S', to discover the acceleration of this frame, provided that the direction of the true vertical were already known. However, a greater interest attaches to learning about the acceleration through dynamic methods. That is the concern of the next section. ACCELERATED FRAMES AND INERTIAL FORCES From what has bcen said, it is clear that inertial frames have a very special status. AU inertial frames are eguivalent in the sense that it is impossible by means of dynamical experiments to dis- cover their motions in any absolute sense— only their relative motions are significant. Out of this dynamical equivalence comes what is called the Newtonian principle of relativity: There is no dynamical observation that leads us to prefer one inertial frame to another. Hence, no dynamical experiment will tell us whether we have a constant velocity through space. As we have just seen, however, a relative acceleration between two frames is dynamically detectable. As observed in accelerating frames, objects have unexpected accelerations. It follows at once, since Newton's law establishes a link between force and ac- celeration, that we have a quantitative basis for calculating the magnitude of the inertial force associated with a measured acceleration. Conversely, and more importantly, we have a dynamical basis for inferring the magnitude of an acceleration from the inertial force associated with it. This is the underlying 497 Accelerated frames and inertial forees principle of all the instruments known as accelerometers. They function because of the inertial property of some physical mass. To makc the analysis explicit, consider the motion of a particle P with respect to two reference frames like those con- sidered in the last section and shown in Fig. 12-2: an inertial frame S and an accelcrated frame S'. We then have, once again, x = X 1 + x, y = y' The velocity components of P as measured in the two frames are thus given by where v s = dx s /dt at any particular instant. If S' has a constant acccleration a, we can put d s = v + at, but the condition of constant acceleration is not at all necessary to our analysis. Taking the time derivatives of the instantaneous velocity components, we then get a z = a' x + a, Oy = a' y where a, is the instantaneous acceleration of the frame S'. Although we have chosen to introduce the calculation in terms of Cartesian components, it is clear that a single vector statement relates the acceleration a of P, as measured in S, to its accelera- tion a' as measured in S' together with the acceleration a, of S' itself: a = a' (12-1) Multiplying Eq. (12-1) throughout by m, we recognize the left- hand side as giving the real (net) foree, F, that is aeting on the particle, since this defines the true cause of its acceleration as measured in an inertial frame. That is, in the S frame, F = «n (12-2) but, using Eq. (12-1), this gives us F = »ra' + ma, (12-3) We now come to the crucial question: How do we interpret Eq. (12-3) from the standpoint of observations made within the 498 Inertial forees and noninertial frames accelerated frame S' itself? Newton's viewpoint — that the net force on an object is the cause of accelerated motion (F nct = ma)- is so deeply ingrained in our thinking that we are strongly motivated to preserve this relationship at all times. When we observe an object accelerating, we interpret this as due to the action of a net force on the object. Can we achieve a mathematical format that has the appearance °f Fnct = wa f° r tne present case of an accelerated frame of reference? Yes. By transferring all terms but ma' to the left and treating these terms as forces that act on m, and have a resultant F', which is of the correct magnitude to produce just the observed acceleration a' : F' = F - ma, = ma' (12-4) The net force in the S' frame is thus made up of two parts: a "real" force, F, with components F x and F y , and a "fictitious" force equal to —ma s , which has its origin in the fact that the frame of reference itself has the acceleration +a„. An important special case of Eq. (12-4) is that in which the "real" force F is zero, in which case the particle, as observed in S', moves under the action of the inertial force — ma s alone. The result expressed by Eq. (12-4) is not mercly a mathe- matical trick. From the standpoint of an observer in the ac- celerating frame, the inertial force is actually present. If one took steps to keep an object "at rest" in S', by tying it down with springs, these springs would be observed to elongate or contract in such a way as to provide a counteracting force to balance the inertial force. To describe such a force as "fictitious" is therefore somewhat misleading. One would like to have some convenient label that distinguishes inertial forces from forces that arise from truc physical interactions, and the term "pseudo- force" is oftcn used. Even this, however, does not do justice to such forces as experienced by someone who is actually in the accelerating frame. Probably the original, strictly technical name, "inertial force," which is free of any questionable over- tones, remains the best description. As an illustration of the way in which the same dynamical situation may be described from the different standpoints of an inertial frame, on the one hand, and an accelerated frame, on the other, consider a simple pendulum suspended from the roof of a car. The mass of the bob is m. In applying F = ma from the standpoint of a frame of reference S attached to the earth 499 Accelerated l'rames and inertial forces Fig. 12-4 Forces acting on a suspended mass in (a) a slation- ary car, (b) a car moving al consianl velocity, and (c) a car imdergoing a posiliue acceleration. (assumed nonrotating), one can draw isolation diagrams for the possible motions of the car as shown in Fig. 12-4. In each case, there are just two (real) forces acting on the bob: F„, the foree of gravity, and T, the tension in the string. Cases (a) and (b) do not involve acceleration and the application of F = ma is trivial. In (c), the bob undcrgoes acceleration toward the right and the string hangs at an angle with some increase in its tension (from rto 7",). The isolation diagram of Fig. 12-5 (a) leads us to apply F = ma as follows: Horizontal component: 7"i sin0 = ma Vertical component : Ty cos 6 — mg = In the S' frame, however, because of the acceleration of the frame, there will be an additional foree of magnitude ma in the direetion opposite to the acceleration of the frame. Figure 12-5(b) shows an isolation diagram for the bob as seen in S'. The bob is in equilibrium. Here, application of F' = ma' gives (because a' = 0): Fig. 12-5 Forces on an object thai is al resi relaliue lo an acceleraled car (a) as judged in an inerlial frame, and (b) as judged in the acceler- aled frame. 500 Inerlial forces and noninertial frames Ti sin S - ma = O 7"i cos 8 — mg = Thus the equilibrium inclination of the pendulum is defined by the condition tan 9 = - g (12-5) ACCELEROMETERS Ffe. /2-5 (a) Tie hanging in equilib- rium within an accelerated cehicle. (h) Quantitatwe ac- celerometer based on measuring the eguilib- rium angle of a simple plumbline. (c) Car- penter's level (in Ihis case a pivoted marker immersed in liquid of greater densily) can be used as an acceler- omeler. (d) A bubble irapped in a curved tube of liguid gices direct readings of acceleralion. Tliis form of accelerometer was deuised by W. V. Walton (Edueation Research Center, M.I.T.). Figure 12-7 shows an example of its use. The result expressed by Eq. (12-5) provides the theoretical basis for a simple accelerometer. If we have first established the true vertical direction, representing e = 0, the observation of the angle of inclination of a pendulum at any subsequent time tells us the value of a through the equation a = g tan d For example, if a passenger in an airplane lets his tie, or a key- chain, hang freely from his fingers during the takeoff run, he can make a rough estimate of the acceleration, which is usually almost constant [Fig. 12-6(a)]. If he also records the time from the beginning of the run to the instant of takeoff, he can obtain 501 Accelerometer* 10 20 30 40 50 60 70 80 90 100 110 120 130 140 /, sec Fig. 12-7 (a) Record obtaimd with the accelerometer ofFig. 12-6(rf) before and afler takeoffofa commercial jet aircraft. The accelerometer was held se- as to record the horizontal component of acceleration only. Not e the sharp decrease in a at takeoff. (b) Graph ofvelocity versus time, obtained by graphical integration of(a). (c) Graph of distance versus time, obtained by graphical integration of(b). a fairly good estimate of the length of the run and the takeoff speed. If he is more ambitious, he can go armed with a card, as in Fig. 12-6(b), already marked out as a goniometer (= angle measurer) or even directly calibrated in terms of acceleration measured in convenient units (e.g., mph per second). ' Another simple accelerometer is obtainable readymade in the form of a carpenter's level made of a small pivoted float that is completely immersed in a liquid [Fig. 12-6(c)]. AH these devices make use of the fact that the natural direction of a plumbline in an ac- celerated frame is defined by the combination of the gravitational acceleration vector g and the negative of the acceleration a of the frame itself. A quite sensitive accelerometer of this same basic type, with the further advantages of a quick response and a quick attain- ment of equilibrium (without much overshoot or oscillation) can be made by curving a piece of plastic tubing into a circular arc and filling it with water or acetone until only a small bubble remains [see Fig. 12-6(a)]. Figure 12-7 (a) shows the record of acceleration versus time as obtained with such an accelerometer during the takeoff of a jet aircraft. Figures 12-7(b) and (c) show the results of numerically integrating this record so as to obtain the speed and the total distance traveled. Accelerometers of a vastly more sophisticated kind can be made by using very sensitive strain gauges, with electrical mea- suring techniques, to record in minute detail the deformations of elastic systems to which a mass is attached. Figure 12-8 shows in schematic form the design of such an instrument. If the object on which the accelerometer is mounted undergoes an acceleration, the inertial force experienced by the pendulum bob begins to deflect it. This, however, unbalances slightly an elec- trical capacitance bridge in which the pendulum forms part of two of the capacitors, as shown. An error signal is obtained which is used both to provide a measure of the acceleration and to drive a coil that applies a restoring force to the pendulum. Such an accelerometer unit may have a useful range from about 10 -5 "g" to more than 10 g. 'A book entitled Science for the Alrplane Passenger by Elizabeth A. Wood (Ballantine, New York, 1969) has such a goniometer on its back cover and a discussion of its use in the text. The book also describes a host of other ways in which airplane passengers can discover or apply scientific principles during their travels. 503 Accelerometers Integrating circuit Signal generator Fig. 12-8 Electro- mechanical acceler- omeler system. Amplifier Bridge unbalance signal ACCELERATING FRAMES AND GRAVITY In all our discussions of accelerated frames, we have assumed that the observers know "which way is up" — i.e., they know the direction and magnitude of the force of gravity and treat it (as we have done) as a real force, whose source is the gravitating mass of the earth. But suppose our frame of reference to be a completely enclosed room with no access to the external sur- roundings. What can one then deduce about gravity and inertial forces through dynamical experiments wholly within the room? We shall suppose once again that there is an observer in a frame, S, attachcd to the earth. This observer is not isolated; he is able to verify that the downward acceleration of a particle dropped from rest is along a line perpendicular to the earth's surface and hence is directed toward the center of the earth. ' He is able to draw the orthodox conclusion that this acceleration is due to the gravitational attraction from the Iarge mass of the earth. Our second observer is shut up in a room that defines the frame S'. Initially it is known that the fioor of his room is ■We are slill ignoring the rolation of the earth, which causes this statement to be not quite correct. A falling object does nol fail exactly parallel to a plumb- line. We shall come back to this when we diseuss rotating reference frames. 504 Incrlial 101 •ccs and non inertial frames' horizontal and that its walls are vertical. In subsequent mea- suremcnts, however, the observer in S' finds that a plumbline hangs at an angle to what he had previously taken to be the vertical, and that objects dropped from rest travel parallel to his plumbline. The observers in S and S' report their findings to one another by radio. The observer in S' then concludes that he has three alternative ways of accounting for the component of force, parallel to the floor, that is now exerted on all particles as observed in his frame: 1 . In addition to the gravitational force, there is an inertial force in the — x direction due to the acceleration of his frame in the +x direction. 2. His frame is not accelerating, but a large massive object has been set down in the — x direction outside his closed room, thus excrting an additional gravitational force on all masses in his frame. 3. His room has been tilted through an angle e and an extra mass has been placed beneath the room to increase the net gravitational force. (This is close to being just a variant of alternative 2.) In supposing that all three hypotheses work equally well to explain what happens in S', we must assume that the additional massive object, postulated in alternatives 2 and 3, produccs an effectively uniform gravitational field throughout the room. From dynamical experiments made entirely within the closed room, there is no way to distinguish among these hypotheses. The acceleration of the frame of reference produces effects that are identical to those of gravitational attraction. Inertial and gravitational forces are both proportional to the mass of the object under examination. The procedures for detecting and measuring them are identical. Moreover, they are both de- scribablc i n tcrms of the properties of a field (an acceleration field) that has a certain strength and direction at any given point. An object placed in this field experiences a certain force without benefit of any contaet with its surroundings. Is all this just an interesting parallel, or does it have a deeper significance? Einstein, after pondering these questions, concluded that there was indecd somcthing fundamental here. In particular, the completely exact proportionality (as far as could be determincd) between gravitational force and inertial mass suggested to him that no physical distinetion could be drawn, at least within a 505 Accelerating frames and eravity (a) (b) Fig. 12-9 (a) Apple falling inside a box t ha I rests on Ihe earth. (b) Indistinguishable molion when Ihe apple is inside an acceleraled box i n ouler space. limited region, between a gravitational field and a general ac- celeration of the reference frame (see Fig. 12-9). He announced this— his famous principle of equivalence— in 191 1. 1 The propor- tionality of gravitational force to inertial mass now becomes an exact necessity, not a n empirical and inevitably approximate result. It is also implied that anything traversing a gravitational field must follow a curved path, because such a curvature would appear on purely kinematic and geometrical grounds if we re- placed the gravitational field by the equivalent acceleration of our own reference frame. I n particular, this should happen with rays of light (see Fig. 12-10). With the help of these ideas Einstein proceeded to construct his general theory of relativity, which (as we pointed out in Chaptcr 8) is primarily a geometrical theory of gravitation. Fig. 12-10 Successive stages in ihe path of a hori- zontally Iraveling object as observed within an enclosure acceleraling certically upward. This illustrates ihe eauicalence ofgraviiy and a general acceleration of the reference frame. 'A. Einstein, Ann. Phys. (4) 35, 898 (1911), reprinted in translated form in The Principle of Relaticiiy (W. Perrett and G. B. Jeffery, translators), Methuen, London, 1923 and Dover, New York, 1958. 506 Inertial forces and noninertial frames CENTRIFUGAL FORCE We shall now consider a particular kind of inertial force that always appears if the motion of a particle is described and analyzed from the standpoint of a rotating reference frame. This force — the centrifugal force — is familiar to us as the force with which, for example, an object appears to puli on us if we whirl it around at the end of a string. ' To introduce it, we shall consider a situation of just this kind. Suppose that a "tether ball" is being whirled around in horizontal circular motion with constant speed (Fig. 12-11). We shall analyze the motion of the ball as seen from two view- points: a stationary frame S, and a rotating frame S' that rotates with the same (constant) rotational speed as the ball. For con- venience, we align the coordinate systems with their z and z' axes (as well as origins) coincident. The rotational speed of S' relative to S will be designated w (in rad/sec). Figure 12-11 shows the analysis with respect to these two frames. The essential con- clusions are these: 1. From the standpoint of the stationary (inertial) frame, the ball has an acceleration (— w 2 r) toward the axis of rotation. The force, F r , to cause this acceleration is supplied by the tether- ing cord, and we must have (In S) F, = —mo>-r 2. From the standpoint of a frame that rotates so as to kcep exact pace with the ball, the acceleration of the ball is zero. We can maintain the validity of Newton's law in the rotating frame if, in addition to the force F r , the ball experiences an inertial force Fi, equal and opposite to F r , and so direeted radially outward : ^{«-A + ft-O [Fi = m<i> 2 r The force Fi is then what we call the centrifugal force. The magnitude of the centrifugal force can be cstablishcd experimentally by an observer in the rotating frame S'. Let him hold a mass m stationary (as secn in his rotating frame) by •The name "centrifugal" comes from ihe Latin: cenlrum, the center, and fugere, to ffee. 507 Centrifugal force Procedure Viewed from stationary frame S Viewed from rotating frame i" l *\gj A \^ r . — a — -*— -S Pictorial sketch of h ' - <& problem Bali is observed to move with speed u in a circle of radius r (angular speed u<) Boy turns around at the same angular speed o> as the ball; from his point of view, the ball is at rest "Isolate fhe body." Draw all forces that act on the ball \ T \>1 , nng T = tension in cord mg = force of gravity F, = inertial force due to viewing the problem from a rotating frame For ease of calculation we resolve forces into T sin $ \ T cos e 'mg r sin 9 T cos e F, components in mutually perpendicu- lar directions 1 1 mg Vertical Direction Vertical Direction We now analyze the problem in terms of F = ma Because there is no vertical ac- celeration, we conclude that the net vertical force must be zero; hence T cos - mg Because there is no vertical acceleration, we conclude that the net vertical force must be zero; hence; T cos n = mg Horizontal Direction Horizontal Direction The object is t' therefore acc force (i.e.. th forces) is hor center of the be equal in n hence T sin e - This force is di inward aveling in a circle. elerating: the net B sum of all three zontal toward the circle, and must lagnitude to mv*lr; rected radially The object is "at rest," therefore the sum of all the forces on it must be zero; hence F, is equal in magnitude to T sin A From the analysis in the left column. it is given by ,. mv 2 . t i = — — = m<u'r and is directed outward. We call F, the centrifugal force Fig. 12-11 Mol ion of a suspended ball, which is traveling in a horizontal circle, as analyzed from the earth's reference frame and from a frame rotating with the ball. 508 Inertial forces and noninerlial frames Spring balance fastened to the axis of rotation, which is perpendicular to the plane of the paper- Fig. 12-12 Measurement of the W \^1>1)00000 t>~- A-. force needed to hold an object at rest in a rotating reference frame. attaching it to a spring balance (Fig. 12-12). If the mass is at any location except on the axis of rotation, the spring balance will show that it is exerting on the mass an inward force propor- tional to m and r. If the observer in S' is informed that his frame is rotating at the rate of w rad/sec, he can confirm that this force is equal to mwV. The observer explains the extension of the spring by saying that it is counteracting the outward centrifugal force on m which is present in the rotating frame. Furthermore, if the spring breaks, then the net force on the mass is just the centri- fugal force and the object will at that instant have an outward acceleration of u 2 r in response to this so-called "fictitious" force. Once again the inertial force is "there" by every eriterion we can apply (except our inability to find another physical system as its source). The magnitude of the centrifugal force is given, as we have seen, by the equation A Fccntrifugai = m— = mw 2 r (radially outward) (12-6) A nice example of our almost intuitive use of this force, under conditions in which there is nothing to balance it, is provided by situations such as the following: We have been washing a piece of straight tubing, and we want to get it dry on the inside. As a first step we get rid of the larger drops of water that are sitting on the inside walls. And we do this, not by shaking the tube longitudinally, but by whirling it in a circular arc [Fig. 12-13(a)]. The analysis of what happens as we begin this rotation gives a particularly elear picture of the difference between the deseriptions of the process in stationary and rotating frames. It also provides us with a different way of deriving the formula for the centrifugal force itself. Suppose that a drop, of mass m, is sitting on the inner wali of the tube at a point A [Fig. 12-13(b)], a distance r from the axis of rotation. Assume that the tube is very smooth, so that 509 Centrifugal force Fig. 12-13 (a) Shak- ing a drop ofwater oul ofa tube. (A) Analysis of initial motion in lerms of centrifugal forces. the drop encountcrs no resistance if it moves along the tube. The drop must, however, be carried along in any transverse movement of the tube resulting from the rotation. Then if the tube is suddenly set into motion and rotated through a small angle A0, the drop, receiving an impulse normal to the wali of the tube at A, moves along the straight line AC. This, however, means that it is now further from the axis of rotation than if it had been fixcd to the tube and had traveled along the circular arc AB. We have, in fact, BC = r sec A0 - r Now sec AS = (cosA0)-' « [1 - KAfl) 2 ]" 1 « 1 + KA<0 2 Therefore, BC ~ £r(A0) 2 We can, however, exprcss A0 in terms of the angular velocity w and the time Al: A0 = u A/. Thus we have BC = iw 2 r(Af) 2 This is then recognizable as the radial displacement that occurs in time Ar under an acceleration u 2 r. Hence we can put "ccntrifiiciil — °> r and so fcentrifuRal = "'W 2 /" 510 inertial forces and noninertial frames CENTRIFUGES Notice, then, that what is, in fact, a small transverse displace- ment in a straight line, with no real force in the radial direction, appears in the frame of the tube as a small, purely radial dis- placement under an unbalanced centrifugal force. The physical fact that the drop i s moved outward along the tube is readily understood in terms of either description. (We should add, how- ever, that our analysis as it stands does only apply to the initial step of the motion. Once the drop has acquired an appreciable radial velocity, things become more complicated.) The term "centrifugal force" is frequently used incorrectly. For example, one may read such statements as "The satellite does not fail down as it moves around the earth because the centrifugal force just counteracts the force of gravity and hence there is no net force to make it fail." Any such statement flouts Newton's first law — A body with no net force on it travels in a straight line. . . . For if the satellite is described as moving in a curved path around the earth, it must also have an unbalanced force on it. The only frame in which the centrifugal force does balance the gravitational force is the frame in which the satellite appears not to move at all. One can, of course, consider the description of such motions with respect to a reference frame rotating at some arbitrary rate different from that of the orbiting object itself. In this case, however, the centrifugal contribution to the inertial forces represents only a part of the story, and the simple balancing of "real" and centrifugal forces does not apply. In particular, let us reemphasize that in a nonrotating frame of reference there is no such thing as centrifugal force. The long- standing confusion that leads people to use the term "centrifugal force" incorrectly has driven at least one author to extreme vexation. In an otherwise sober and quite formal text the author writes: "There is no answer to these people. Some of them are good citizens. They vote the ticket of the party that is responsible for the prosperity of the country; they belong to the only true church; they subscribe to the Red Cross drive — but they have no place in the Temple of Science; they profane it." ' The laboratory centrifuge represents an immensely important and direct application of the dynamical principle of centrifugal force. The basic arrangement of a simple type of centrifuge is 'W. F. Osgood, Mechanics, Macmillan, New York, 1937. 511 C?cntrifuacs Fig. 12-14 (a) Ver- lical section through a simple centrifuge. (b) Analysis of radial sedimenlation in terms of cenirifugal forces. shown in Fig. 12-14(a). Carefully balanced tubes of liquid are suspended on smooth pivots from a rotor. When the rotor is made to spin at high speed, the tubes swing upward and outward into almost horizontal positions and may be maintained in this orientation for many hours on end. At any point P in one of the tubes [Fig. 12— 14(b)], distance r from the axis of rotation, there is an effective gravitational field of magnitude w 2 r, which may be made very much greater than g. For example, if r = 15 cm and the rotor spins at 25 rps (co = 50* sec -1 ), the value of w 2 r is about 4000m/sec 2 or 400 g. Small particles in suspension in the liquid will be driven toward the outward (bottom) end of the tube much more quickly than they would ever be under the action of gravity alone. The basis for calculating the drift speed is the formula for the resistive force to motion through a fluid at low speeds, which we first met in Chapter 5. For a spherical particle of radius r and speed v, this force is proportional to the product ro. If the medium is water, the approximate magnitude of the force is given by R(v) « 0.02rv where R is in newtons, r in meters, and v in m/sec. A steady value of v is attained when this force just balances the driving force associated with the effective gravitational field strength, g'. In calculating this driving force it is important to allow for buoyancy effects — i.e., Archimedes' principle. If the density of the particle is p p and the density of the liquid is p ( , the driving force is given by F=y(p„-p,)rV 512 Inertial forces and noninertial (rames This can be more simply expressed if we introduce the true mass m of the particle (= 4irp p r 3 /3), in which case we can put To take a specific example, suppose that we have an aqueous suspension of bacterial particles of radius 1 p., each with a mass of about 5 X 10 -15 kg and a density about 1.1 times that of water. If w6 take for g' the value 400 g calculated earlier, we find F« 2 X 10~ 12 N We thus obtain a drift speed given by 2 X 10~ 12 ]n _ 4 , ^ 2X10-^10-» a '° m/S6C This represents a settling rate of several centimeters per hour, which makes for effective separation in reasonable times, whereas under the normal gravity force alone one would have only a millimeter or two per day. The above example represents what one may regard as a more or less routine type of centrifugation, but in 1925 the Swedish chemist T. Svedberg opened up a whole new field of research when, by achieving centrifugal fields thousands of times stronger than g, he succccded in measuring the molecular weights of proteins by studying their radial sedimentation. The type of machine he developed for this purpose was appropriately named the ultracenlrifuge, and Svedberg succeeded in producing centri- fugal fields as high as about 50,000 g. The physicist J. W. Beams has taken the technique even further through his development of magnetic suspensions, in vacuum, that dispense with me- chanical bearings altogether. The rotor simply spins in empty space, with carefully controlled magnetic fields to hold it at a constant vertical level against the normal puli of gravity. By such methods Beams has produced centrifugal fields equivalcnt to about 10° g in a usable centrifuge and fields as high as 10 9 # in tiny objects (e.g., spheres of 0.001-in. diameter). The limita- tion is set by the bursting speed of the rotor; this defines a maxi- mum value of u proportional to l/r (see Chapter 7, p. 208). Since the centrifugal field g' is equal to u 2 r, and the limiting speed sets an upper limit to ur, it may be seen that the attainable value of g' varies as l/r. The techniquc of ultracentrifuge methods has been brought 513 Centrifuees to an extraordinary pitch of refinement. It has become possible to determine molecular weights to a precision of bettcr than 1% over a range from about 10 8 (virus particles) down to as low as about 50. The possibitity of measuring the very low molecular weights by this method is particularly impressive. Beams has pointed out that in a solution of sucrose in water, the calculatcd rate of descent of an individual sucrose molecule, of mass about 340 amu and radius about 5 A, would (according to the kind of analysis we gave earlier) be less than 1 mm in 100 years under normal gravity. (A rate as slow as this becomes in fact mean- ingless because, as Beams points out, it would be completely swamped by random thermal motions.) If a field of 10 5 g is available, howevcr, the time constant of the sedimentation process is reduced to the order of 1 day or less, which brings the measurement well within the range of possibility. ! This whole subject of centrifuges and centrifugation is a particularly good application of the concept of inertial force, because the phenomena are so appropriately described in terms of static or quasistatic equilibrium in the rotating frame. CORIOLIS FORCES We have seen how the centrifugal force, mo> 2 r, exerted on a particle of a givcn mass m in a frame rotating at a given angular velocity w, depends only on the distance r of the particle from the axis of rotation. In general, however, another inertial force appears in a rotating frame. This is the Coriolis force, 2 and it depends only on the velocity of the particle (not on its position). We shall introduce this force in a simple way for some specific situations. Later, by introducing vector expressions for rotational motion, we shall develop a succinct notation that gives both the centrifugal and Coriolis forces in a form valid in three dimensions using any type of coordinate system. The need to introduce the Coriolis force is easily shown by comparing the straight-line motion of a particle in an inertial frame S with the motion of the same particle as seen in a rotating frame S'. ■For further reading on this extremely interesting subject, see T. Svedberg and K. O. Pedersen, The Ulrracentrifuge, Oxford University Press, New York, 1960, and J. W. Beams, "High Centrifugal Fields," Physics Teacher 1, 103 (1963). 2 G. Coriolis, J. de VEcole Polytechnique, Cahier 24, 142 (1835). 514 Inertial forces and nonincrtial irames Suppose that S' is a coordinate system attached to a hori- zontal circular table that rotates with constant angular speed w. Let the vertical axis of rotation define the z' axis and suppose that the table surface (in the x'y' plane) has no frietion. A string fastened to the origin holds a partiele on the y' coordinate axis at a radial distance r' a from the axis of rotation. Thus, in the S' frame, the partiele is at rest in equilibrium under the combined forees of the tension in the string and the centrifugal foree. (The vertical foree of gravity and the normal foree of the table surface always add to zero and need not concern us further.) The same partiele is viewed from an inertial frame S which coincides with S' at f = 0. In this stationary frame, the partiele travels with uniform speed ve = ov" in a cirele of constant radius r (= r' ) under the single unbalanced foree of the tension in the string. There is, of course, no centrifugal foree in this inertial frame. At / = the string breaks. In S the partiele then travels Fig. 12-15 Two different deseriptions of the molion of an object that is initially tethered on a rotating disk and begins molion under no forees at t = 0. 0, the string breaks. The dots indicate the positions of the partiele at successive equal time intervals 0, 1, 2, 3 At the same time instants the rotating y' axis has the positions shownby(O), (1), (2), (3),.... As seen in the rotating S' frame The partiele is initially at rest. After the string breaks, it begins to move radially outward, but as soon as it acquires some speed, it veers toward the right of the y' axis. 515 Coriolis forees in a straight line with constant speed v = u>r Q as shown in Fig. 12-15(a). To find the motion in S', we compare in the stationary frame the positions of the y' axis and the corresponding locations of the particle at successive equally-spaced times. We discover that the particle, as observed in S', not only moves radially outward, but also moves farther and farther to the right of the single radial line formed by the rotating / axis. This result is plotted in Fig. 12— 15(b). To explain this motion as observed in the rotating frame, it is necessary to postulate, in addition to the centrifugal force, a sideways deflecting force. This deflecting force is the Coriolis force. In the course of the following dis- cussion, we shall determine its magnitude and show that it always acts at right anglcs to any velocity v' in the S' frame. We can find the magnitude of the Coriolis force by in- vestigating another simple motion in these two frames. Suppose that, instead of the situation just described, we make a particle follow a radially outward path in the rotating frame at constant velocity v' r . In this frame there must be no net force on the particle. Hence we shall have to supply some real (inward) force to counteraet the varying (outward) centrifugal force as the particle moves. We shall not concern ourselves with these radial components but will concentrate our attention only on the transverse-foree components. In this way we can remove from consideration the distortion of the trajectory by the centrifugal force, which is purely radial. How does the motion appear in the two frames? Figure 12— I6(b) shows the straight-line path of the object in the rotating frame. But the path of the object in the stationary frame is a curved line AB as shown in Fig. 12-16(a). In S the transverse velocity vg (= ur) is greater at B than at A, because the radial distance from the axis is greater at B. Hence there must be a real transverse force to produce this increase of velocity seen in the stationary frame. This real force might be provided, for example, by a spring balance. What does this motion look like in the rotating frame? In S' the object moves outward with constant speed and hence has no acceleration [see Fig. 12-16(b)]. This means, as we have said, that there can be no net force on the object in the rotating frame. But since an observer in S' sees the spring balance exert- ing a real sideways force on the object in the +6 direetion, he infers that there is a counteraeting inertial force in the — diree- tion to balance it. This is the Coriolis force. 516 Inertial forees and noninertial frames Fig. 12-16 (a) Lab- oratory view of the path of a particle that moves radially out- ward on a rotaling table. (b) The motion as it appears in the rotaling frame itself. _____^_____^_____________^__ As seen in the stationary s frame The mass point foilows the curved path frorrr A to 8. As seen in the rotating S' frame The mass point moves radially outward with constant speed. co(r + 8r) To determine its magnitude, let O A and OB in Fig. 12-17 be successive positions of the same radial line at times separated by A/. Let OC be the bisector of the angle A0. The velocity perpendicular to OC changes by the amount Avg during At, where Av e = [co(r + Ar) cos(A0/2) + v T sin(A0/2)] - [tar cos(A0/2) - v, sin(A0/2)] For small angles we can put the cosine equal to 1 and the sine equal to the angle, which leads to the following very simple expression for Ao$ : Avb a» coAr + v r A6 The transverse acceleration ao is thus given by a 6 = ai(Ar/Al) + u r (A0/Al) But Ar/Al = c r and A6/At = o> Hence at = 2uv, Ft = 2mwo r Fig. 12-17 Basis of calculaling the Coriolis force for a particle mocing radially at constant speed with re- spect to a rotating tahle. 517 Coriolis forces This gives us the real force needed to cause the real acceleration as judged in S. But as observed in the rotating frame S', there is no acceleration and no net force. Hence the existence of the Coriolis force, equal to — 2mcov' T , is inferred. (Note that v T — Vr.) This inertial force is i n the negative B' direction, opposite to the spring force, and is at right angles to the direction of motion of the particle : Fs'(Coriolis) = -2mwo' r (12-7) An important feature, which you should verify for yourself, is that if we had considered a radially inward motion (v' r negative), then we would have inferred the existence of a Coriolis force acting in the positive B' direction. In both cases, therefore, the Coriolis force acts to deflect the object in the same way with respect to the direction of the velocity v' itself— to the right if the frame S' is rotating counterclockwise, as we have assumed, or to the left if S' rotates clockwise. It turns out, in fact, as we shall prove later, that even in the case of motion in an arbitrary direction the Coriolis force is always a deflecting force, exerted at right angles to the direction of motion as observed in the rotating frame. The Coriolis force is very real from the viewpoint of the rotating frame of reference. If you want toconvince yourself of the reality of this "fictitious" force, ride a rotating merry-go- round and try walking a radial line outward or inward. (Proceed cautiously — the Coriolis force is so unexpected and surprising that it is easy to lose one's balance!) DYNAMICS ON A MERRY-GO-ROUND As we have just mentioned, the behavior of objects in motion within a rotating reference frame can run strongly counter to one's intuitions. It is not too hard to get used to the existence of the centrifugal force acting on an object at rest with respect to the rotating frame, but the combination of centrifugal and Coriolis effects that appear when the object is set in motion can be quitc bewildering, and somctimes entertaining. Suppose, for example, that a man stands at point A on & merry-go-round [Fig. 12-18(a)] and tries to throw a ball to someone at B (or perhaps aims for the bull's-eye of a dart board placed there). Then the thrown object mystcriously veers to the right and misses its target every time. One can blame part of this, of course, on the centrifugal force itself. Howevcr, it is to be noted that 518 Inertial forccs and nonincrlial framcs Fig. 12-18 (a) Tra- jectories ofobjects as lliey appear to ob- servers on a rotating lable. (b) An object projected on afriction- less rotating table can return to its starting point. since the magnitude of the centrifugal force is mu 2 r and that of the Coriolis force is 2mwv', the ratio of these two forces is pro- portional to v'fur. Thus if v' is made much greater than the actual peripheral speed of the merry-go-round, the peculiarities of the motion are governed almost entirely by the Coriolis effects. If this condition holds, the net deflection of a moving object will always be to the right with respect to v' on a merry-go-round rotating counterclockwise. Thus if the positions A and B in Fig. 12-18(a) are occupied by two people trying to throw a ball back and forth, each will have to aim to the left in order to make a good throw. An extrcme case of this kind of behavior can cause an object to follow a continuously curved path that brings it back to its starting point, although it is not subjected to any real forces at all. This phcnomenon has been demonstrated in the highly entertaining and instructive film, Frames of Reference. 1 A dry-ice puck, launched at point A on a tabletop of plate glass [Fig. 12-18(b)], can be caused by a skilled operator to follow a trajectory of the kind indicated. GENERAL ECjUATION OF MOTION IN A ROTATING FRAME 2 The goal of this discussion will be to relate the time derivatives of the displacement of a moving object as observed in a sta- '" Frames of Reference," by J. N. P. Humc and D. G. Ivey, Education Develop- ment Center, Newton, Mass., 1960. 2 This section may be omitted by a reader who is willing to take on trust its final results — that the total inertial force in a rotating frame is the combination of the centrifugal force with a Coriolis force corresponding to a generalized formof Eq. (12-7). 519 General equation of motion in a rotating frame Fig. 12-19 Use of angular velocity as a vector to define the linear velocity ofa parlicle on a roiating lable: v = to X r. tionary frame S' and in a rotating frame 5'. To set the stage, we shall introduce the idca that angular velocity may be represented as a vector. Consider first a point P on a rotating disk [Fig. 12-19(a)]. It has a purely tangential velocity, v$, in a direction at right angles to the radius OP. We can describe this velocity, in both magnitude and direction, if we define a vector according to the same convention that we introduced for torque in Chapter 4. That is, if the fingcrs of the right hand are curled around in the sense of rotation, keeping the thumb extended as shown in the figure, then w is represented as a vector, of length propor- tional to the angular speed, in the direction in which the thumb points. Thus with to pointing along the positive z direction, one is defining a rotation that carries each point such as P from the positive x direction toward the positive y direction. The rotation of the disk is in this case counterclockwise as seen from above. The velocity of P is now given by the vector (cross) produet of to with the radius vector r: v = to X r (12-8 This veetor-produet exprcssion is valid in three dimensions also, if the position vector r of P is measured from any point on the axis of rotation, as shown in Fig. 12-19(b). The radius of the cirele in which P moves is R = r sin e. Thus we have v = vg = cor sin 6, in a direction perpendicular to the plane defined by to and r. That is precisely what Eq. (12-8) gives us. Next, we consider how the change of any vector during a 520 Inertial forees and nonincrtial frames Fig. 12-20 (a) Change of a vector, analyzed in terms ofits change as measured on a rotaling lable, to- gel/ier with the change I due lo rolalion of the lable iiself. (6) Sim- ilar analysisfor an arbitrary veclor referred to any origin on the axis ofrotation. small time interval At can be exprcssed as the vector sum of two contributions: 1. The change that would occur if it were simply a vector of constant length embedded in the rotating frame S'. 2. The further change described by its change of length and direction as observed in S'. In Fig. 12-20(a) we show this analysis for motion confined to a plane. The vector A at time / is represented by the line CD. If it remains fixed with respect to a rotating table, its direction at time t + At is given by the line CE, where A8 = w At. Thus its change due to the rotation alone would be represented by DE, where DE = A Ad = Au At. From the standpoint of frame S' this change would not be observed. There might, how- ever, be a change represented by the line EF; we shall denote this as AA S — the change of A as observed in S'. The vector sum of DE and EF, i. e., the line DF, then represents the true change of A as observed in 5. We therefore denote this as AX S . In Fig. 12-20(b) we show the corresponding analysis for three dimensions. The length of DE is now equal to A sin 6 Aip; its direction is perpendicular to the plane defined by w and A. Since A<p = u At, we can put vector displacement DE = (u X A) A/ The displacement A A s- may be in any direction with respect to DE, but the two again combine to give a net displacement DF which is to be identified with A As. Thus we have 521 General eqnalion of motion in a rotating frame AA S = AA S ' + (w X A)Af We can at once proceed from this to a relation between the ratcs of change of A as observed in S and S', respectively: \drjs \ di J s- + « X A (12-9) This is a very powerful relation because A can be any vector we please. First, we shall choose A to be the position vector r. Then (d\/dt)s is the true velocity, v, as observed in S, and (dA/dt)g> is the apparent velocity, v', as observed in S'. Thus we immediately have v = v' + « X r (12-10) Next, we shall choose A to be the velocity v: Now (dv/dt)s is the true acceleration, a, as observed in S. The quantity (d\/dt) s - is, however, a sort of hybrid — it is the rate of change in S' of the velocity as observed in S. We can make more sense of this if we substitute for v from Eq. (12-10); we then have ®, -(& + •*$„ The two terms on the right of this equation are now quite recog- nizable; (d\'/dl)s- is the acceleration, a', as observed in S', and (dr/dt)s> is just v'. Thus we have (£), = •*- X v' Substituting this in Eq. (12-11) we thus get a = a'+«Xv' + «Xv We do not need to have both v and v' on the right-hand side, and we shall again substitute for v from Eq. (12-10). This gives us finally a = a' + 2u X v' + u X (w X r) (12-12) A remark is in order regarding the last term, which involves 522 Inerlial fbrees and noninertial Frames the cross product of three vectors. According to the rules of vector algebra, the cross product inside the parentheses is to be taken first, then the other cross product performed. A nonzero answer will result for all cases where the angle formed by u and r is other than 0° or 180°. Performing the cross products in the reverse (incorrect) order, however, would result in zero for all cases, regardless of the angle between these vectors. Multiplying Eq. (12-12) throughout by the mass m of the object, we recognize the left side as the net external force on the mass as seen in the stationary system. ma = F nct = m»' + 2m(w X v') + m[u X (<o X r)] In the rotating frame of reference, the object m has the accelera- tion a'. We may preserve the format of Newton's second law in this accelerated frame of reference by rearranging the above equation, so as to be able to write Fi* = roa' where (12-13a) FL = F net - 2m(« X v') - m[u X (to X r)] "real" Coriolis centrifugal (12-13b) force force force inertial forces The mathematical form of Eq. (12-13b) shows that both the Coriolis force and the centrifugal force are in a direction at right angles to the axis of rotation defined by u. The centrifugal force, in particular, is always radially outward from the axis, as is clear if one considers the geometrical relationships of the vectors involved in the product — u X (w X r), as shown in Fig. 12-21. The equation also shows that the Coriolis force es Fig. 12-21 Relalion of the vectors in- volved informing the centrifugal accelera- tion — c» X (w X r). uxr -» x (u x r) 523 General equation of motion in a rotating frame would reverse if thc direction of u were reversed, but the direction of the centrifugal force would remain unchanged. The specification of F' in Eq. (12-13) can be made entirely on the basis of measurements of position, velocity, and accelera- tion as observed within the rotating fratne itself. The centrifugal term, involving the vector r, might seem to contradict this, but we could just as well put r' instead of r, because observers in the two frames do agree on the vector position of a moving object at a given instant, granted that they use the same choice of origin. To summarize, we have established by the above calculation that the dynamics of motion as observed in a uniformly rotating frame of reference may be analyzed in terms of the following three categories of forces: "Real": F nnt Coriolis: -2m(oi X v') This is the sum of all the "real" forces on the object such as forces of contaet, tensions in strings, the force of gravity, electrical forces, magnetic forces, and so on. Only these forces are seen in a stationary frame of reference. The Coriolis force is a deflecting force always at right angles to the velocity v' of the mass m. If the object has no velocity in the rotating frame of reference, there is no Coriolis force. It is an inertial force nol seen in a stationary frame of reference. Note minus sign. The centrifugal force depends on position only and is always radially outward. It is an < inertial force not seen in a stationary frame I of reference. We could equally well write it (as — m[o> X (<•> X r')]. Note the minus sign. THE EARTH AS A ROTATING REFERENCE FRAME In this seetion we shall consider a few examples of the way in which the earth's rotation affects the dynamical processes oc- curring on it. The local value of g If a partiele P is at rest at latitude X near the earth's surface, then as judged in the earth's frame it is subjected to the gravita- 524 Inertial forces and noninertial frames Fig. 12-22 (a) Forces on an objecl at rest Ott the earth, as interpreted in a reference frame that roiates wilh l/ie eart/i. (b) An objecl falling from rest relalive to Ihe earth undergoes an eastward displacement. (c) The fall- ing motion of (b), as seen from a frame thal does nol roiale wilh ihe earth. tional force F„ and the centrifugal force F ccnt shown in Fig. 12-22(a). The magnitude of the latter is given, according to Eq. (12-13b), by the equation ^cent = rnu> 2 R sin 9 = mu R cosX where R is the earth's radius. We have already discussed in Chapter 8 the way in which this centrifugal term reduces the local magnitude of g and also modifies the Iocal direction of the vertical as dcfined by a plumbline. The analysis is in fact much simpler and clearer from the standpoint of our natural reference frame as defined by the earth itself. We have, as Fig. 12-22(a) shows, the following relations: F a — moi Rcos X Fr = F, - Fccnt COS X F$ = Fcnt sin X = mu R sin X cos X Deviation offreelyfalling objects (12-14) If a particle is releascd from rest at a point such as P in Fig. 12-22(a), it begins to accelerate downward under the action of a net force F' whose components are given by Eq. (12-14). As 525 The earth as a rotating reference frame soon as it has any appreciable velocity, however, it also expe- riences a Coriolis force given by the equation Fcorioiu = -2mn X v' (12-15) Now the velocity v' is i n the plane PON containing the earth's axis. The Coriolis force must be perpendicular to this plane, and a consideration of the actual directions of u and v' shows that it is eastward. Thus if we set up a local coordinate system defined by the local plumbline vertical and the local easterly direction, as in Fig. 12-22(b), the falling object deviates eastward from a plumbline AB and hits the ground at a point C. The effect is very small but has been detected and measured in careful ex- periments (see Problem 12-24). To calculate what the deflection should be for an object falling from a given height h, we use the fact that the value of v' to be inserted in Eq. (12-15) is extremely well approximated by the simple equation of free vertical fail : v' = gt where v' is measured as positive downward. Thus if we label the eastward direction as x', we have m— r-r- = (2mu> cos X)gt Integrating this twice with respect to t, we have x' = $gwt a cos X (12-16) For a total distance of vertical fail equal to h, we have t = (2h/g) U2 , which thus gives us X ,_2V|«C<|X A 3,2 (121?) Inserting approximate numcrical values (w = 2xday -1 ~7X 10 -5 sec -1 ), one finds x' ~ 2 X 10- 5 A 3/2 cos X (x' and h in m) Thus, for example, with h = 50 m at latitude 45°, one has x" « 5 mm, or about ? in. It is perhaps worth reminding oneself that the effects of inertial forces can always be calculated, if one wishes, from the 526 Inertial forces and noninertial frames standpoint of an inertial frame in which these forces simply do not exist. In the present case, one can begin by recognizing that a particle held at a distance h above the ground has a higher eastward velocity than a point on the ground below. For sim- plicity, let us consider how this operates at the equator (X = 0). Figure I2-22(c) shows the trajectory of the falling object as seen in a nonrotating frame. The object has an initial horizontal velocity given by DO, = o>(R + h) After a time t it has traveled a horizontal distance x given, very nearly, by coRt. With the object now at P (see the figure) the gravitational force acting on it has a very small component in the negative x direction. We have, in fact, F x *» — jf ( « —mgmt Hence A Integrating once, we have dx , 2 ■j t ~ "o*- is°» Substituting the value v Qx = w(R + h), this gives, as a very good approximation, d f t = co(« + /o - iga 2 Integrating a second time, we have x - «(* + h)i - \gut 3 However, the point O at the earth's surface is also moving, with a constant speed of coR. Thus, when the falling object hits the ground at C, the point O has reached O', where 00' = uRl. Hence we have x' = O'C w oj/>/ - i*cor 3 If we substitute h = \gt 2 , we at once obtain the result given by 527 The cari h as a rotating reference frame Fig. J 2-23 Format ion ofa cyclone In the northern hemisphere, under the aclion of Coriolis forces on the moving air masses. Low Eq. (12-16) for X = 0. [Or, of course, we can substitute t = V2h/g and arrive at Eq. (12-17)]. Patterns of atmospheric circulation Because of the Coriolis effect, air masses being driven radially inward toward a low-pressure region, or outward away from a high-pressure region, are also subject to deflecting forces. This causes most cyclones to be in a counterclockwise direction in the northern hemisphere and clockwise in the southern hemisphere. The origin of these preferred rotational directions may be seen in Fig. 12-23, which shows the motions of air in the northern hemisphere moving toward a region of low pressure. The hori- zontal components of the Coriolis force deflect these motions Fig. 12-24 Tiros satellite photo- graph of a cyclone. (Courtesy of Charles W. C. Rogers and N.A.S.A.) i BJty.\~ -— 31 I ^fl ^^^-*" ^ f^m *•*-»• ^^^B^^ |§£$ . RJ— «.'— fl_ ■C^K2l^ V' - ^frfp-1 528 Inertial forces and noninerlial frames toward the right. Thus, as the air masses converge on the center of the low-pressure region, they produce a net counterclockwise rotation. For air moving north or south over the earth's surface the Coriolis force is due east or due west, parallel to the earth's surface. If we consider a 1-kg mass of air at a wind velocity of 10 m/sec (about 22 mph) at 45° north latitude, a direct applica- tion of Eq. (12-15) gives us , Fcorioii, - 2mui/ sin X « (2)(1)(2tt X lO" 5 ) X (10)(0.707) w 10" 3 N If we had considered air flowing in from east or west, the Coriolis forces would not be parallel to the earth's surface, but their components parallel to the surface would be given by the same equation as that used above. (Verify this.) The approximate radius of curvature of the resultant motion may be obtained from 2 R or 2 R = m — ^ « 1 X -r^r-r. = 10' 5 m (about 60 miles) rCorioli. 10 •» As air masses move over hundreds of miles on the earth's sur- face, they often form huge vortices — as is dramatically shown in the Tiros weather satellite photograph in Fig. 12-24. Occasionally one reads that water draining out of a basin also circulates in a preferred direction because of the Coriolis force. In most cases, the Coriolis force on the flowing water is negligible compared with other larger forces which are present; however, if extremely precise and careful expcriments are per- formed, the effect cari be demonstrated. 1 The Foucault pendulum No account of Coriolis forces would be complete without some mention of the famous pendulum experiment named after the French physicist J. B. L. Foucault, who first demonstrated in 1851 how the slow rotation of the plane of vibration of a pen- 'See, for example, the film "Balhtub Vortex," an exccrpt from "Vorticity," by A. H. Shapiro, National Council on Fluid Mechanics, 1962. 529 The earth as a rotating reference frame Fig. 12-25 (a) A pendulum swinging along a north-south line at lalilude X. (b) Palh of pendulum bob, as seenfrom above. (The change ofdireclion per swing is, however, grossly exaggerated.) dulum could be used as evidence of the earth's own rotation. It is easy, but rather too glib, to say that of course we are simply seeing the effect of the earth turning beneath the pendulum. This description might properly be used for a pendulum sus- pended at the north or south pole. One can even press things a little further and say that at a given latitude, X [see Fig. 12-25(a)] the earth's angular velocity vector has a component w sin X along the local vertical. This would indeed lead to the correct result— that the plane of the pendulum rotates at a rate corresponding to one complete rotation in a time 7"(X) given by HX) = lir >sinX 24 esc X hours (12-18) But the pendulum is, after all, connected to the earth via its suspending wire, and both the tension in the wire and the gravita- tional force on the bob lie in the vertical plane in which the pendulum is first set swinging. (So, too, is the air resistance, if this needs to be considered.) It is the Coriolis force that can be invoked to give a more cxplicit basis for the rotation. For a pendulum swinging in the northern hemisphere, the Coriolis force acts always to curve the path of the swinging bob to the right, as indicatcd in exaggerated form in Fig. 12-25(b). As with the Coriolis force on moving air, the effect does not depend on the direction of swing— contrary to the intuition most of us probably have that the rotation is likely to be more marked when the pendulum swings along a north-south line than whcn it swings east-west. 530 Inertial forces and noninertial frames THE TIDES Fig. 12-26 (a) Double tidal bulge as it would be if the earlh's rotation did not dis- place it. The size of the bulge is enor- mously exaggerated. (b) Approximate true orientation ofthe tidal bulges, carried eastwardby the earth's rotation. As everyone knows, the production of ocean tides is basically the consequence of the gravitational action of the moon — and, to a lesser extent, the sun. Thus we could have discussed this as an example of universal gravitation in Chapter 8. The analysis of the phenomenon is, however, considerably helped by intro- ducing the concept of inertial forces as developed in the present chapter. The feature that probably causes the most puzzlement when one first learns about the tides is the fact that there are, at most places on the earth's surface, two high tides every day rather than just one. This corresponds to the fact that, at any instant, the general distribution of ocean levels around the earth has two bulges. On the simple model that we shall discuss, these bulges would be highest at the places on the earth's surface nearest to and farthest from the moon [Fig. 12-26(a)]. While the earth performs its rotation during 24 hr, the positions of the bulges would remain almost stationary, being defined by the almost constant position of the moon. Thus, if one could imagine the earth completely girdled by water, the depth of the water as measured from a point fixed to the earth's solid surface would pass through two maxima and two minima in each revolution. A better approximation to the observed facts is obtained by con- sidering the bulges to be dragged eastward by friction from the land and the ocean floor, so that their equilibrium positions with respect to the moon are more nearly as indicated in Fig. 12-26(b). To conclude these preliminary remarks, we may point out that the bulges are, in fact, also being carried slowly eastward all the time by the moon's own motion around the earth. This motion (one complete orbit relative to the fixed stars every 531 The tides 27.3 days) has the consequence that it takes more than 24 hr for a given point on the earth to make successive passages past a particular tidal bulge. Specifically, this causes the theoretical time interval between successive high tides at a given place to be close to 12 hr 25 min instead of precisely 12 hr (see Problem 2-15). For example, if a high tide is observed to occur at 4 p.m. one day, its counterpart next day would be expected to occur at about 4:50 p.m. Now let us consider the dynamical situation. The first point to appreciate is the manner in which the earth as a whole is being accelerated toward the moon by virtue of the gravita- tional attraction between them. With respect to the CM of the earth-moon system (inside the earth, at about 3000 miles from the earth's center), the earth's center of mass has an acceleration of magnitude a c given by Newton's laws: MrOc = GMeM„, r m 2 i.e., ac = GM m (12-19) where M m and r„, are the moon's mass and distance. What may not be immediately apparent is that every point in the earth receives this same acceleration from the moon's attraction. If Flg. 12-27 The or- bital motion ofthe earth about the moon does not by itselfin- volce any rotation of the earth; the line AiBi is carried into the parallel configura- tion A2B2. 532 Inertial forces and noninertial frames one draws a sketch, as shown in Fig. 12-27, of the arcs along which the earth's center and the moon travel in a certain span of time, one is tempted to think of the earth-moon system as a kind of rigid dumbbell that rotates as a unit about the center of mass, O. It is true that the moon, for its part, does move so that it presents always the same face toward the earth, but with the earth itself things are different. If the earth were not rotating on its axis, every point on it would follow a circular arc identical in size and direction to the arc C X C 2 traced out by the earth's center. The line AiBy would be translated into the parallel line A 2 B 2 . The earth's intrinsic rotation about its axis is simply superposed on this general displacement and the associated acceleration. This is where noninertial frames come into the picture. The dynamical consequences of the earth's orbital motion around the CM of the earth-moon system can be correctly described in terms of an inertial force, —mac, experienced by a particle of mass m wherever it may be, in or on the earth. This force is then added to all the other forces that may be acting on the particle. In the model that we are using — corresponding to what is called the equilibrium theory of the tides — the water around the earth simply moves until it attains an equilibrium configuration that remains stationary from the viewpoint of an observer on the moon. Now we know that for a particle at the earth's center, the centrifugal force and the moon's gravitational attraction are equal and opposite. If, however, we consider a particle on the earth's surface at the nearest point to the moon [point A in Fig. 12-28(a)], the gravitational force on it is greater than the centrifugal force by an amount that we shall call/ : Fig. 12-28 (a) Dif- ference belween cen- trifugal force and the earth's gravity at the points nearest to and farthest from the moon. (b) Tide- producing force at an arbitrary point P, showing existence of a transverse component. 533 The tides GM m m _ GM m m Since R E « r m (R E « r m /60), we can approximate this expres- sion as follows: /o = i.e., GM m m [(-r-] /„«H™^*, (12 _ 20) By an exactly similar calculation, we find that the tide-producing force on a particle of mass m at the farthest point from the moon [point B in Fig. 12-28(a)] is equal to —f ; hence we recognize the tendency for the water to be pulled or pushed away from a midplane drawn through the earth's center (see the figure). By going just a little further we can get a much better in- sight into the problem. Consider now a particle of water at an arbitrary point P [Fig. 12-28(b)]. Relative to the earth's center, C, it has coordinates (x, y), with x = R E cos 6, y = R E sin 6. The tidal force on it in the x direction is given by a calculation just like those above: m IGM^n x _ IGM^n ^ cqs g (]2 _ 21) This yiclds the results already obtained for the points A and B if we put d = or ir. In addition to this force parallel to the line joining the centers of the earth and the moon there is also, however, a transverse force, because the line from P to the moon's center makes a small angle, a, with the x axis, and the net gravita- tional force, GM m m/r' 2 , has a small component perpendicular to x, given by GM m m . , ... , /„ = j— sin a (with r = r*) Now we have y tana = Since a is a very small angle [<tan ■ (/?/.;//•„,), which is about 1°] wc can safely approximate the above exprcssion: 534 Inertial forces and noninerlial frames Fig. 12-29 Pattern of tide-producing forces around the earlh. The circular dashed line shows where the undisturbed waler surface would be. y_ _ R E sin d r m r- The component/„ of the tidal force is then given by GM m m GM m m f v » 5— y = 5— Re sin (12-22) We see that this transverse force is greatest at d = ir/2, at which point it is equal to half the maximum value (/ ) of f x . Using Eqs. (12-21) and (12-22) together, we can develop an over-all picture of the tide-producing forces, as shown in Fig. 12-29. This shows much more convincingly how the forces act in such directions as to cause the water to flow and redistribute itself in the manner already qualitatively described. TIDAL HEIGHTS; EFFECT OF THE SUN 1 How high ought the equilibrium tidal bulge to be? If you are familiar with actual tidal variations you may be surprised at the 'This section goes well beyond the scope of the chapter as a whole but is added for the interest that it may have. 535 Tidal heights; eflect of the sial result. The equilibrium tide would be a rise and fail of less than 2 ft. We can calculate this by considering that the work done by the tidal force in moving a particle of water from D to A (Fig. 12-29) is equivalent to the increase of gravitational po- tential energy needed to raise the water through a height h against the earth's normal gravitational puli. ' The distance h is the difference of water levels between A and D. Now, using Eqs. (12-21) and (12-22) we have dW = f z dx + f„dy GM m m ,_ , . = ir~ (2xdx - ydy) W r r R E r° GM m m Wda = 2x dx — / ydy r m 3 U o Jr b 3GM m m 2 Setting this amount of work equal to the gain of gravitational potential energy, mgh, we have _ 3GM m R B 2 (12 _ 23) 2gr m a The numerical values of the relevant quantities are as follows: G = 6.67 X 10~ n m 3 /kg-sec 2 M m = 7.34 X 10 22 kg r m = 3.84 X 10 8 m R E = 6.37 X 10° m g = 9.80 m/sec 2 Substituting these in Eq. (12-23) we find A « 0.54 m « 21 in. The great excess over this calculated value in many places (by factors of 10 or even more) can only be explained by considering the problem in detailed dynamical terms, in which the accumula- tion of water in narrow estuaries, and resonance effects, can completely alter the scale of the phenomenon. The value that we have calculated should bc approximated in the open sea. The last point that we shall consider here is the effect of the sun. Its mass and distance are as follows: ■Technically, this condition corrcsponds to the water surface being an energy equipotential. 536 Incrlial forces and noninertial irames M, = 1.99 X 10 30 kg r„ = 1.49X10" m If we directly compare the gravitational forces exerted by the sun and the moon on a particle on the earth, we discover that the sun wins by a large factor: F, MJr? M.(r m \\ F„ M m /r m 2~ Mm \r.) ° What matters, however, for tide production is the amount by which these forces change from point to point over the earth. This is expressed in terms of the gradient of the gravitational force: • m = 2p /=AF=-^Ar (,2-24) Putting M = M m , r = r m , and Ar = ±R E , we obtain the forces ±/ corresponding to Eq. (12-20). We now see that the comparative tide-producing forces due to the sun and the moon are given, according to Eq. (12-24), by the following ratio: Substituting the numerical values, one finds fy « 0.465 Jm This means that the tide-raising ability of the moon exceeds that of the sun by a factor of about 2.15. The effects of the two combine linearly— and, of course, vectorially, depending on the relative angular positions of the moon and the sun. When they are on the same line through the earth (whether on the same side or on opposite sides) there should be a maximum tide equal to 1.465 times that due to the moon alone. This should happen once every 2 weeks, approximately, when the moon is new or full. At intermediate times (half-moon) when the angular posi- tions of sun and moon are separated by 90°, the tidal amplitude should fail to a minimum value equal to 0.535 times that of the moon. The ratio of maximum to minimum values is thus about 2.7. 537 Tidal heights; efl'cct of the sun THE SEARCH FOR A FUNDAMENTAL INERTIAL FRAME The phenomena that we have discussed in this chapter seem to leave us in no doubt that the acceleration of one's frame of reference can be dctected by dynamical means. They suggest that a very special status does indeed attach to inertial frames. But how can we be sure that we have identified a true inertial frame in which Galileo's law of inertia holds exactly? We saw at the very beginning of our discussion of dynamics that the earth itself represents a good approximation to such a frame for many purposes, especially for dynamical phenomena whose scale in distance and time is small. But we have now seen * abundant evidence that a laboratory on the earth's surface is accelerated. If the laboratory is at latitude X (see Fig. 12-30), each point in it is accelerating toward the earth's axis of rotation with an acceleration given by a\ = oi 2 R cos X with 01 = 27r/86,400sec- 1 R = 6.4 X 10 6 m This gives a\ = 3.4 X 10- 2 cosXm/sec 2 This acceleration of a frame of reference tied to the earth is, as we know, not the simplest case of an accelerated frame. The linearly accelerated frames with which we began this chapter are much more readily analyzed. It was, however, the phenomena associated with rotating frames that led Newton to his belief in absolute space and in the absolute character of accelerations. Near the beginning of the Principia he describes a celebrated Fig. 12-30 Acceleration toward ihe earth's axis by virtue ofits rotation. 538 Inertial Corces and noninertiai frames Bucket rotating water stationary Bucket and water rotating together Bucket stationary water rotating Fig. 12-31 Main features ofllie experiment that Newton guolecl as evidence oflhe absolute character of rotation and the associated acceleration. experiment that he made with a bucket of water. It is ari experi- ment that anyone may readily repeat for himself. The bucket is hung on a strongly twisted rope and is then released. There are three key observations, depicted in Fig. 12-31: 1. At first the bucket spins rapidly, but the water remain's almost at rest, before the viscous forces have had time to set it rotating. The water surface is flat, just as it was before the bucket was released. 2. The water and the bucket are rotating together; the water surface has become concave (see Problem 12-18). 3. The bucket is suddenly stopped, but the body of water continues to rotate, and its surface remains curved. ' Clearly, said Newton, the relative motion of the bucket and the water is not the factor that determines the curvature of the water surface. It must be the absolute rotation of the water in space, and its attendant acceleration, that is at the bottom of the phenomenon. And with the help of F = ma, we can account for it quantitatively. Newton's argument is a powerful one. He could point to further evidence in support of his views in the bulging of the earth itself by virtue of its rotation. The equatorial diameter of the earth is greater than the polar diameter by about 1 part in 300. It seems almost obvious, even without detailed calcula- tion, that this is closely tied to the fact that a^/g is about g£t5 at the equator and is zero at the poles (although the detailed cal- culation is, in fact, a bit messy). Newton did not stop here, of course, He held the key of universal gravitation. Even a nonrotating earth would not be an inertial frame, because the whole earth is accelerating toward the sun. For this system we have a> = 2ir/(3.l6 X lO^sec- 1 R = 1.49 X 10 n m a 2 = o) 2 /? = 5.9 X 10- 3 m/sec 2 Newton does not suggest that he actually performed this third step, but it represents a natural completion of the experiment as one might perform it for oneself. 539 The search for a fundamental inertial frame If we could conceive of an object that was immune to the gravita- tional attraction of the sun, it would not obey the law of inertia as observed from a reference frame attached to the earth. From Newton's standpoint the acceleration is real and absolute and is linked to the existence of a well-defined gravitational force pro- vided by the sun. That was about the end of the road as far as Newton was concerned. For him the system of the stars provided the arena in which the motions that he so brilliantly analyzed took place. A reference frame attached to these fixed stars could be taken to constitute a true inertial system, even though it might not coin- cide with the absolute space in which he believed. Today, thanks to the work of astronomers, we know a good deal about the motions of some of those "fixed" stars. We have come to be aware of our involvement in a general rotation of our Galaxy. The sun would appear to be making a complete circuit of the Galaxy in about 2.5 X 10 8 years at a radial distance of about 2.5 X 10 4 light-years from the center. For this motion we would have co~ 2tt/(8 X 10 15 )sec-' «« 2.4 X 10 20 m a 3 « 10- 10 m/sec 2 It looks as though this acceleration can be reasonably accounted for by means of Newton's law of universal gravitation, if we regard the solar system as having a centripetal acceleration under the attraction of all the stars lying within its orbit. But no dynamical experiments that we do on earth require us to take into account this extremely minute effect — or, even, for most purposes, the revolution of the earth about the sun. (The rotation of the earth on its own axis is, however, an important con- sideration — and indeed an important aid in such matters as gyroscopic navigation.) Figure 12-32 schematizes the three rotating frames in which we find ourselves (we ignore here the acceleration caused by the moon). But we still have not found an unaccelerated object to which we can attach our inertial frame of reference. In fact, we could extend this tantalizing search even further. There is some evi- dence that galaxies themselves tend to cluster together in groups containing a few galaxies to perhaps thousands. Our local group consists of about 10 galaxies. Although individual galaxies could have rather complex motions with respect to each other, 540 Inertial force* and non inertial frames Fig. 12-32 Accelera- | tions ofany laboratory reference frame at- tached to the earth's surface. this group is believed to have a more or less common motion through space. So where are the "fixed" stars or other astronomical objects to which we can attach our inertial frame of reference? It ap- pears that referring to the "fued stars" is not a solution and contains an uncomfortable element of metaphysics (although we frequently usc this phrasc as a shorthand designation for the establishment of an inertial frame). This does not mean that the astronomical search for an inertial frame has been without value. For, at lcast up to the galactic level, it would seem that apparent departures from the law of inertia can be traced to identifiable accelerations of the reference frame in which motions are observed. However, the quest is incomplete, and so it seems likely to remain. Ultimatcly, therefore, we rely on an operational 541 Tho search for a fundamental inertial frame definition based upon local dynamical experiments and observa- tion. We define an inertial frame to be one in which, experi- mentally, Galileo's law of inertia holds. The very existence of the inertial property remains, however, a deep and fascinating prob- lem, and we shall end the chapter with a few remarks about this most fundamental feature of dynamics. SPECULATIONS ON THE ORIGIN OF INERTIA Not everyone accepted Newton's view that the phenomena associated with rotating objects demonstrated the absolute character of acceleration. The philosopher-bishop, George Berkeley, was perhaps the first person to argue 1 that all motions, including rotational ones, only have meaning as motions relative to other objects. The circling of two spheres around their center of mass could not, he said, be imagined in a space that was otherwise empty. Only when we introduce the background represented by the stars do we have a basis for recognizing the existence of such motion. About 150 years later (in 1872) the German philosopher Ernst Mach presented the same idea in much more cogent form. He wrote: For mc, only relative motions exist . . . and I can see, in this regard, no distinction between rotation and translation. Ob- viously it does not matter if we think of the earth as turning round on its axis, or at rest while the fixed stars revolve around it . . . But if we think of the earth at rest and the fixed stars revolving around it, there is no flattening of the earth, no Foucault's experiment and so on— at least according to our usual conception of the law of inertia. Now one can solve the difflculty in two ways. Either all motion is absolute, or our law of inertia is wrongly expressed ... I prefer the second way. The law of inertia must be so conceived that exactly the same thing results from the second supposition as from the first. By this it will be evident that in its expression, regard must be paid to the masses of the universe. 2 ■In his iraet De Moiu, written in 1717, 30 years after the publication of Newton's Principia. 2 E. Mach, Hislory and Rool of the Principle of the Consercation of Energy, (2nd ed.), Barth, Leipzig (1909). English translation of the 2nd edition by P. Jourdain, Open Court Publishing Co., London, 1911. Actually the first sentence of the quolation is taken from Mach's classic book, The Science of Mechanics, first published in 1883. 542 Inertial forees and nonincrtial frames Thus was born the profound and novel idea — subsequently to become famous as Mach's principle — that the inertial property of any given object depends upon the presence and the distribu- tion of other masses. Einstein himself accepted this idea and took it as a central principle of cosmology. If one admits the validity of this point of view, then one sees that the whole basis of dynamics is involved. For consider the method that we described in Chapter 9 (p. 319) for finding the ratio of the inertial masses of two objects. This ratio is given as the negative inverse ratio of the accelerations that they produce by their mutual interaction: /Ml _ (22 W2 O i This looks very simple and straightforward, but it is clear that our ability to attach specific values to the individual accelera- tions, as distinct from the total relative acceleration, depends completely on our having identified a reference frame in which these accelerations can be measured. For this purpose the physical background provided by other objects is essential. In looking critically at the phenomena of rotational motion, Mach attacked some intuitive notions that are much more deep- seated than any that we have in connection with straight-line motion. He considered the evidence provided by Newton's rotating-bucket experiment which we discussed in the last section. It is quite clear that the curvature of the water surface is related overwhelmingly to the cxistence of rotation relative to the vast amount of distant matter of the universe. When that relative rotation is stopped, the water surface becomes fiat. When the bucket rotates and the water remains still (both relative to the fixed stars), the shape of the water surface remains unaffected. But, said Mach, that may bc only a matter of degree. "No one," he wrote, "is competent to say how the experiment would turn out if the sides of the vessel increased in thickness and mass until they were ultimately several leagues thick." His own belief was that this rotation of a monster bucket would in fact generate the equivalent of centrifugal forces on the water insidc it, even though this water had no rotational motion in the accepted sense. This is a startling idea indced. Let us prcsent it in a slightly dirTerent context. We know that the act of giving an object an acceleration a, with respcct to the inertial frame defined by the fixed stars, calls into play an inertial force, equal to — /wa, that expresses the resistance of the object to being accelerated. In 543 Spcculations on the origin of inertia Mach's view we are equally entitled (indeed, compelled) to accept a description of the phenomenon in a frame always attached to the object itself. In this frame the rest of the universe has the acceleration a' (= —a) and the inertial force ma' that the object experiences must be ascribable to the acceleration of the other masses. This then brings us to the quantitative question: If a mass M, at distance r, is given the acceleration a relative to a given object, what contribution does it make to the total inertial force ma that the object experiences? Since we know that the force is proportional to m, we can argue on the grounds of symmetry and relativity that it must be proportional to M also. But at this point we entcr a more speculative realm. A very suggestive analogy is provided by electromagnetic interactions. If two electric charges, q t and q it are separated by a distance r, we know that the static force exerted by q x on q 2 is given by _ fr?l<?2 where k is a constant that depends on the particular choice of units. If, however, the charge q x is given the acceleration a there is an additional force that comes into play, directly proportional to a and inversely proportional to the distance: , kq x q 2 a ri2 = 5 — c 2 r where c is the speed of light. Since this force falls off more slowly with distance than the static interaction, it can survive in ap- preciable magnitude at distances at which the static l/r 2 force has become negligible. This is, in fact, the basis of the electro- magnetic radiation field by which signals can be transmitted over large distances. Suppose now that we assume an analogous situation for gravitational interactions. The basic static law of force is known to be GMm The force on m associated with an acceleration of M would then be given by , GMma n . ,,., Fl2 = ~&T ( } 544 Inertial forces and noninerlial framcs On this basis we can estimate the relative magnitudes of the contributions from various masses of interest — the earth, the sun, our own Galaxy, and the rest of the universe. Ali we have to do is to calculate the values of M/r for these objects. The results are shown in Table 12-1, using numbers to the nearest power of 10 only. (The value of M for the universe as a whole is the somewhat speculative value quoted in Chapter 1.) We TABLE 12-1: RELATIVE CONTRIBUTIONS TO INERTIA Source M, kg r, m M/ 'r, kg/m M/r (relative) Earth 10 25 10 7 10 18 10- 8 Sun 10 3O 10 11 10 19 io- 7 Our Galaxy 10* l 10« 10 20 10-° Universe ] 52 10 2G 10 2e 1 see that, according to this theory, the effect of a nearby object, even one as massive as the earth itself, would be negligible com- pared to the effect of the universe at large. The total inertial force called into existence if everything in the universe acquires an acceleration a with respect to a given object would be obtained by summing the forces F' l2 of Eq. (12-26) over all masses other than m itself: _ ^GM «"inertial = ma £_, — — c*r This, however, should be identical with what we know to be the magnitude of the inertial force as directly given by the value of ma. Thus the theory would require the following identity to hold: Ef-1 02-27) universe u ' It is clear from Table 12-1 that even such a large local mass as our Galaxy represents only a minor contribution ; what we are invoived with is a summation over the approximately uniform distribution of matter represented by the universe as a whole. If we regard it as a sphere, centered on ourselves, of mean density p and radius R v (« 10 10 light-years = 10 26 m), we would have univerao ' «'O » 545 Speculations on the oriain of inertia PROBLEMS The total mass is however given by Mu = — pRu Thus we have, on this simple picture (based on Euclidean geometry) E M 3 Mu ,-20 1 / — = - — « 10 kg/m universe ** Using the values G m 10 -10 N-m 2 /kg 2 and c 2 m 10 17 m 2 /sec 2 , we would then have E^-io- 1 f — c&r universe " Taking into account the uncertainties in our knowledge of the distribution of matter throughout space, many would say that the factor of about 10 that separates the above empirical value from the theoretical value (unity) called for by Eq. (12-27) is not significant. The result is intriguing, to say the least, and many cosmologists have accepted as fundamentally correct this develop- ment from the primary ideas espoused by Mach and Einstein. ' 12-1 A single-engine airplane flies horizontally at a constant speed v. In the frame of the aircraft, each tip of the propeller sweeps out a circle of radius R at the rate of n revolutions per second. Obtain an equation for the path of a tip of the propeller as viewed from the earth. 12-2 A person observes the position of a post from the origin of a reference frame (S') rigidly attached to the rim of a merry-go-round, as shown in the figure. The merry-go-round (of radius R) is rotating with angular velocity u, the distance of the post from the axis of the merry-go-round is D, and at / = 0, the coordinates of P in 5' are x' = D - R,y' = (equivalently, r' = D - R, d' = 0). (a) Find the coordinates r'(i), 6'(t) of the post; also give the corresponding *'(/) and y'(t). •For further reading on this fascinating topic, see, for example, R. H. Dicke, "The Many Faces of Mach," in Gracitation and Relalicily (ed. H.-Y. Chin and W. F. Hoffmann, eds.), W. A. Benjamin, New York, 1964; N. R. Hanson, "Newton's First Law," and P. Morrison, "The Physics of the Large," both in Beyond the Edge of Cenainly (R. G. Colodny, ed.), Prentice-Hall, Engle- wood Cliffs, N.J., 1965; D. W. Sciama, The Unity ofihe Unicerse, Doubleday, New York, 1961, and The Physical Foundaiions of General Relalicily, Doubleday, New York, 1969. 546 Inertial forccs and noninertial frames (b) By differentiating the results of (a), obtain the velocity and acceleration of the post in both Cartesian and polar coordinates. (c) Make a plot of the path of the post in S'. 12-3 A boy is riding on a railroad flatcar, on level ground, that has an acceleration a in the direction of its motion. At what angle with the vertical should he toss a ball so that he can catch it without shift- ing his position on the car? 12-4 A railroad train traveling on a straight track at a speed of 20 m/sec begins to slow down uniformly as it enters a station and comes to a stop in 100 m. A suitcase of mass 10 kg having a coefficient of sliding friction n = 0.15 with the train's floor slides down the aisle diiring this deccleration period. (a) What is the acceleration of the suitcase (with respect to the ground) during this time? (b) What is the velocity of the suitcase just as the train comes toahalt? (c) The suitcase continues sliding for a period after the train has stopped. When it comes to rest, how far is it displaced from its original position on the floor of the train? 12-5 A man weighs himself on a spring balance calibrated in newtons which indicates his weight as mg = 700 N. What will he read if he repeats the observation while riding an elevator from the first to the twelfth floors in the following manner? (a) Belween the first and third floors the elevator accelerates at the rate of 2 m/sec 2 . (b) Between the third and tenth floors the elevator travels with the constant velocity of 7 m/sec. (c) Between the tenth and twelfth floors the elevator decelerates at the rate of 2 m/sec 2 . (d) He then makes a similar trip down again. (e) If on another trip the balance reads 500 N, what can you say of his motion? Which way is he moving? 12-6 If the cocHicient of friction between a box and the bed of a truck is /x, what is the maximum acceleration with which the truck can 547 Problems climb a hill, making an angle 6 with the horizontal, without the box's slipping on the truck bed ? 12-7 A block of mass 2 kg rests on a frictionless platform. It is attached to a horizontal spring of spring constant 8 N/m, as shown in the figure. Initially the whole system is stationary, but at t = the platform begins to move to the right with a constant acceleration of 2 m/sec. 2 As a result the block begins to oscillate horizontally relative to the platform. 2kg 8 N/m immfmr- TJ T? 2 m/sec 2 (a) What is the amplitude of the oscillation? (b) At t = 2v/3 sec, by what amount is the spring longer than it was in its initial unstretched condition? 12-8 A plane surface inclined 37° (sin -1 f) from the horizontal is accelerated horizontally to the left (see the figure). The magnitude of the acceleration is gradually increased until a block of mass m, orig- inally at rest with respect to the plane, just starts to slip up the plane. The static friction force at the block-plane surface is characterized by (a) Draw a diagram showing the forces acting on the block, just before it slips, in an inertial frame fbced to the floor. (b) Find the acceleration at which the block begins to slip. (c) Repeat part (a) in the noninertial frame moving along with the block. 12-9 A nervous passenger in an airplane at takeoff removes his tie and lets it hang loosely from his fingers. He observes that during the takeoff run, which lasts 30 sec, the tie makes an angle of 15° with the vertical. What is the speed of the plane at takeoff, and how much runway is needed ? Assume that the runway is level. 12-10 A uniform steel rod (density = 7500 kg/m 3 , ultimate tensile strength 5 X 10 8 N/m 2 ) of length 1 m is accelerated along the direc- tion of its length by a constant force applied to one end and directed away from the center of mass of the rod. What is the maximum allowable acceleration if the rod is not to break? If this acceleration is exceeded, where will the rod break? 12-11 (a) A train slowed with deceleration a. What angle would the liquid level of a bowl of soup in the dining car have made with the horizontal ? A child dropped an apple from a height h and a distance d from the front wali of the dining car. What path did the apple take as observed by the child ? Under what conditions would the apple have hit the ground? The front wali ? 548 Tncrtial forces and noninertial frames */3 tt 3^ (b) As a reward for making the above observations, the parents bought the child a helium-filled balloon at the next stop. For fun, they asked him what would happen to the balloon if the train left the station with acceleration a'. Subsequently, they were surprised to find his predictions corrcct. What did the precocious child answer? 12-12 An elevator has a downward acceleration equal to g/3. Inside the elevator is mounted a pulley, of negligible friction and inertia, over which passes a string carrying two objects, of masses m and 3»i, respectively (see the figure). (a) Calculate the acceleration of the object of mass 3m relatice to the elevator. (b) Calculate the force exerted on the pulley by the rod that joins it to the roof of the elevator. (c) How could an observer, completely isolated inside the elevator, explain the acceleration of m in terms of forces that he him- self could measure with the help of a spring balance? 12-13 In each of the following cases, find the equilibrium position as well as the period of small oscillations of a pendulum of length L: (1) In a train moving with acceleration a on level tracks. (2) In a train free-wheeling on tracks making an angle 6 with the horizontal. (3) In an elevator falling with acceleration a. 12-14 The world record for the 16-lb hammer throw is about 70 m. Assuming that the hammer is whirled around in a circle of radius about 2 m before being let fly, estimate the magnitude of the puli that the thrower must be able to withstand. 12-15 (a) A man rides in an elevator with vertical acceleration a. He swings a bucket of water in a vertical circle of radius R. With what angular velocity must he swing the bucket so that no water spills? (b) With what angular frequency must the bucket be swung if the man is on a train with horizontal acceleration a? (The plane of the circle is again vertical and contains the direetion of the train's acceleration.) 12-16 Consider a thin rod of material of density p rotating with con- stant angular velocity a> about an axis perpendicular to its length. (a) Show that i f the rod is to have a constant stress 5 (tensile force per unit area of cross seetion) along its length, the cross-sectional area must deercase exponentialIy with the square of the distance from the axis: — frr a 2 A = Aue where k = po>~/2S [Consider a small segment of the rod between r and r -f Ar, having a mass im = pA(r) Ar, and notice that the differcnce in tensions at its ends is AT = A(SA).] 549 Problem s (b) What is ihe maximum angular velocity o) max in terms of p, 5 m ax, and A:? (c) The ultimate tcnsilc strength of steel is about 10 9 N/m 2 . Estimate the maximum possible number of rpm of a steel rotor for which the "taper constant" k = 100 m~ 2 (p = 7500 kg/m 3 ). 12-17 A spherically shaped influenza virus particle, of mass 6 X 10 -16 g and diameter 10 -5 cm, is in a water suspension in an ultra- centrifuge. It is 4 cm from the vertical axis of rotation, and the speed of rotation is 10 3 rps. The density of the virus particle is 1.1 times that of water. (a) From the standpoint of a rcference frame rotating with the centrifuge, what is the effective value of 'V ? (b) Again from the standpoint of the rotating reference frame, what is the net centrifugal force acting on the virus particle? (c) Because of this centrifugal force, the particle moves radially outward at a small speed o. The motion is resisted by a viscous force given by F rcs = Jnrr\vd, where d is the diameter of the particle and t\ is the viscosity of water, equal to 10~- cgs units (g/cm/sec). What is cl (d) Describe the situation from the standpoint of an inertial frame attached to the laboratory. [In (b) and (c), account must be taken of buoyancy effects. Think of the ordinary hydrostatics problem of a body completely immersed in a fluid of different density.] 12-18 (a) Show that the equilibrium form of the surface of a rotating body of liquid is parabolic (or, strictly, a paraboloid of revolution). This problem is most simply considered from the standpoint of the rotating frame, given that a liquid cannot withstand forees tangential to its surface and will tend toward a configuration in which such forees disappear. It is instructive to consider the situation from the stand- point of an inertial frame also. (b) It has been proposed that a parabolic mirror for an astro- nomical teleseope might be formed from a rotating pool of mercury. What rate of rotation (rpm) would make a mirror of focal length 20 m? 12-19 To a first approximation, an object released anywhere within an orbiting spacecraft will remain in the same place relative to the spacecraft. More accurately, however, it experiences a net force proportional to its distance from the center of mass of the spacecraft. This force, as measured in the noninertial frame of the craft, arises from the small variations in both the gravitational force and the centrifugal force due to the change of distance from the earth's center. Obtain an expression for this force as a funetion of the mass, m, of the object, its distance AR from the center of the spacecraft, the radius R of the spacecraft's orbit around the earth, and the gravitational ac- celeration gR at the distance R from the earth's center. 550 Inertial forees and noninertial frames 12-20 A circular platform of radius 5 m rotates with an angular velocity w = 0.2 rad/sec. A man of mass 100 kg walks with constant velocity v' = 1 m/sec along a diameter of the platform. At time t = he crosses the center and at time t = 5 sec he jumps off the edge of the platform. (a) Draw a graph of the centrifugal force felt by the man as a function of time in the interval / = to t = 5 sec. (b) Draw a similar graph of the Coriolis force. For both diagrams, give the correct vertical scale (in newtons). (c) Show on a sketch the direction of these forces, assuming the platform to rotate in a clockwise direction as seen from above. 12-21 On a long-playing record (33 rpm, 12 in.) an insect starts to crawl toward the rim. Assume that the coefficient of friction between its legs and the record is 0.1. Does it reach the edge by crawling or otherwise ? 12-22 A child sits on the ground near a rotating merry-go-round. With respect to a reference frame attached to the earth the child has no acceleration (accept this as being approximately true) and ex- periences no force. With respect to polar coordinates fixed to the merry-go-round, with origin at its center: (a) What is the motion of the child? (b) What is his acceleration? (c) Account for this acceleration, as measured in the rotating frame, in terms of the centrifugal and Coriolis forces judged to be acting on the child. 12-23 The text (p. 516) derives the Coriolis force in the transverse (8) direction by considering the motion of an object along a radial line in the rotating frame. Correspondingly, if one considers an object that is moving transversely in the rotating frame, one can obtain the net radial force due to Coriolis and centrifugal effects. Consider a particle on a frictionless turntable rotating with angular velocity w. The particle is initially at rest relative to the turntable, at a distance r from the axis of rotation. (a) Set up a fixed coordinate system S with axes x transversely and y radially and with its origin O at the position of the particle at t = (see the figure). Set up another coordinate system S', with origin O' and axes x' and y', which rotates with the turntable and which coincides with S at / = 0. Show that at a later time / the co- ordinates of a given point as measured in S' and S are related by the following equations, where 8 = wt: x' = x cos 8 + y sin 8 + r sin 8 y' = y cos 8 — x sin 8 — r(l — cos 8) (b) Suppose that, at / = 0, the particle is given a velocity v' 551 Problcms relative to O' in the x' direction. Its subsequent motion will be along the x direction at the constant velocity o' — tar relative to O. Use this to obtain its coordinates x' and / at a later time /. (c) Making the approximations for the case wt « 1, show that for small values of t one can put y' ~ %a' r i 2 , where a' T = co 2 r — Tmvf , This corresponds to the required combination of centrifugal and Coriolis accelerations. (d) If you are feeling ambitious, apply the same kind of analysis for an initial velocity in an arbitrary direction. 12-24 In an article entitled "Do Objects fail South?" [Phys. Rev., 16, 246 (1903)], Edwin Hall reported the results of nearly 1000 trials in which he allowed an object to fail through a vertical distance of 23 m at Cambridge, Mass. (lat. 42° N). He found, on the average, an eastward deflection of 0.149 cm and a southerly deflection of 0.0045 cm. (a) Comparc the easterly deflection with what would be ex- pected from Eq. (12-17). (b) Consider the fact that the development of an eastward component of motion relative to the earth would indeed lead in turn to a southerly component of Coriolis force. Without attempting any detailed analysis, estimate the order of magnitudc of the ratio of the resulting southerly deflection to the predominant easterly deflection. Do you think that an explanation of HalPs results on southerly de- flection can be achieved in these terms? 12-25 Calculate the Coriolis acceleration of a satellite in a circular polar orbit as observed by someone on the rotating earth. Obtain the direction of this acceleration throughout the orbit, thereby explaining why the satellite always passes through the poles even though it is subjected to the Coriolis force. Is there a similar force on a satellite in an equatorial orbit? 12-26 Imagine that a frictionless horizontal table, circular in shape and of radius R, is fitted with a perfectly elastic rim, and that a dry-ice 552 Inertial forces and nonincrtial frames puck is launchcd from a point on the rim toward the center. The puck bounccs back and forth across the table at constant speed v, but because of the Coriolis force it does not quite follow a straight-line path along a diameter. Consider the rate at which the path of the puck gradually turns with respect to the table, and compare the result with that for a Foucault pendulum at the same latitude, X. 12-27 In the text (p. 536) the height of the equilibrium tide is cal- culated by considering the work done by the tide-producing force in carrying a particle of water from point D to point A (see the figure). By considering the work from D to an intermediate point P, one can obtain a general expression for the elevation or depression h{6) of the water at an arbitrary point, relative to what the water level would be in the absence of the tide-producing force. The calculation involves two parts, as follows: (a) Evaluate the work integral of the tide-producing force from D(x = 0, y = Re) to P(x = Re cos 0, y = Re sin d) for a particle of water of mass m. Equating this to the difference of gravitational potential energies, mg(iip — ho), one gets an expression for the dif- ference hp — ho- (b) The total volume of water is a constant. Hence, if ho represents the water depth in the absence of the tide-producing force, we must havc L r/ 2 2TR K -[h(6) - ho]sinddd = Putting the results of (a) and (b) together, you should be able to verify that the deviation of the water level from its undisturbed state is proportional to 3 cos 2 — 1 . 553 Problem s What makes planet s go around the sun ? At the time of Kepler some people answered this problem by saying that there were angels behind them beating their wings and pushing the planet s around an orbit. As you will see, the answer is not very far front the truth. The only difference is that the angels sit in a different direction and their wings push inwards. R. P. feynman, The Character of Physical Law (1965) 13 Motion under central forces we have already seen, especially in Chapter 8, how the motion of objects under the action of forces directed toward some well- defined center is one of the richest areas of study in mechanics. Twice in the history of physics the analysis of such motions has been linked with fundamental advances in our understanding of nature — through the explanation of planetary motions, on the macroscopic scale, and through Rutherford's studies of alpha- particle scattering, which gave man his first clear view of the subatomic world. Up until this point we have limited ourselves to the study of circular orbits, and it is remarkable how much can be learned on that basis. But now we shall begin a more general analysis of motion under the action of central forces. BASIC FEATURES OF THE PROBLEM As we saw in Chapter 1 1 (p. 444), a central force field that is also conservative must be spherically symmetric, and some of the most important fields in nature (notably electrical and gravitational) are precisely of this type. The frequent occurrence of spherically symmetric models to describe physical reality is closely linked to the basic assumption that space is isotropic and is the intuitively natural starting point in building theoretical models of various kinds of dynamical systems. We shall begin with the specific problem of the motion of a single particle of mass w in a spherically symmetric central field 555 Fig. 13-1 (a) Unit veciors associated with radial and trans- verse directions in a plane polar coordinale system. (b) Radial and transverse com- ponents ofan ele- mentary vector displacement Ar. of force. Initially, at least, we shall assume that the object re- sponsible for this central ficld is so massive that it can be regarded as a fixed center that defines a convenient origin of coordinates for the analysis of the motion. The first thing to notice is that the path of the moving par- ticle will lie in a fixed plane that passes through the center of force. This plane is defined by the initial velocity vector v of the particle and the initial vector position r of the particle with re- spect to the center of force. Since the force acting on the particle is in this plane, and sincc thcrc is no component of initial velocity perpendicular to it, the motion must remain confined to this plane of r and v . To analyze the motion we must first pick an appropriate coordinate system. Because the force F is a function of the scalar distance r only and is along the line of the vector r (positively or negatively), it is clearly most convenient to work with the plane polar coordinates (r, 0), as indicated in Fig. 13-l(a). This means that we shall be making use of the acceleration vector expressed in these coordinates. In Chapter 3 (p. 108) we calcu- lated this vector for the particular case of circular motion (r = constant). Now we shall develop the more general expres- sion that embraces changes of both r and 0. Using the unit vectors e r and e» as indicated in Fig. 13-1 (a) we have r = re, dr dr dd V= d, = dt C ' + r di e ° (13-1) (13-2) This equation for v is readily constructed by recognizing that a general infinitesimal change in position, Ar, is obtained by com- bining a radial displacement of length Ar (at constant 0) and a 556 Motion under cenlral forccs transverse displacement of length r A6 (at constant r), as indicated in Fig. 13-1 (b). Alternatively, one can just differentiate both sides of Eq. (13-1) with respect to time, remembering that d(e r )/dt = {dO/dtytt [see Eq. (3-17a)]. We now proceed to differentiate both sides of Eq. (13-2) with respect to / : dv d 2 r , dr d , dr dO , <f$ a = d t = dT* er + d t d, e ' + d l d- t ee + r dT2 e ' , dd d + r dtdt* Substituting d(e r )/dt = (d6/dt)e e , and d(e$)/dt = - (dd/dt)c T , the expression for a can be rewritten as follows: ir. dfi e r + ^1+ 2— — r dfi di dt Cfl (13-3) It will be convenient to extract from this the separate radial and transverse components of the total acceleration : d\ (de\ 2 a ' = dfi- r \j t ) d 2 d , , dr dB at=r-rT + 2-r-r dfi dt dt (13-4) (13-5) The statement of Newton's law in plane polar coordinates can then be made i n terms of these separate acceleration components: F T = m F» = m d\ dfi 1 d^£ dr dd r dfi + dt dt (13-6) (13-7) The above two equations provide a basis for the solution of any problem of motion in a plane, referred to an origin of polar coordinates. We shall, however, consider their application to central forccs in particular. THE LAW OF EQUAL AREAS In the case of any kind of conservative central force, we have F T = F(r) simply, and F e = 0. The second of these immediately implies that a s = 0. Substituting the specific expression for a s 557 The law of equal areas Fig. 13-2 Illustrating the basis of calculating areal velocity (area swept oul per unit lime by ihe radius vector). from Eq. (13-5), we have r£+2$£-0 03-8) dt 2 dt dt This equation contains a somewhat veiled statement of a simple geometrical result — that the vector r sweeps out area at a constant rate. One way of seeing this is to multiply Eq. (13-8) throughout by r: 2d 2 6 drdd ' dfi + lr 7iJi = Q The left-hand side may then be recognized as the derivative with respect to / of the product r 2 dd/dt: rf/s«\ 2d2 drdd dt\ di) r dt2 + ' dt dt If we integrate this expression, we therefore have r 2 ? = const. (13-9) dt Now in Fig. 1 3-2 we show the area AA (shaded) swept out by r in a short time Al. It is the triangle POQ (we take PQ to be indistinguishable from a straight line if it is short enough) and we have AA - iKr + Ar) sin A6 The rate at which area is being swept out, instantaneously, is the limit of AA/At for At —* 0. Since, as we approach this limit, Ar/r — » and sin Ad — > A0, we arrive at the result d A = \ r ^l (,3-10a) dt 2 dt Thus we recognize the constant on the right-hand side of Eq. 558 Motion undcr ccntral f'orces Fig. 13-3 (a) A por- tion of Newton' s manuscript, De Motu, showing the basis of his proofofthe law of equal areas for a cenlral force. (6) En- larged copy of New- ton' s diagram. (.Front J. Herivel, The Back- ground to Newton's Principia, Oxford University Press, London, 1965.) (a) (13-9) as twice the rate (a constanl rate) at which the radius vector r sweeps out area, and we therefore have (Any central force) d A 1 2 dd — = s r — = const. dt 2 t/r (13-10b) The result expressed by Eq. (13-10b) was first discovered by Kepler in his analysis of planetary motions (of which more later). It was stated by him in what is known as his second law (although it was actually the first chronologically). Newton understood it on the samc dynamical grounds as we have discussed above, i.e., as a feature of the motion of an object acted on by any kind of force that is directed always to the same point. He 559 The law of cqunl areas visualized the action of such a force as a succession of small kicks or impulses, which in the limit would go over into a continuously applied infiuence. He set out this view of the process in a tract written in 1684 (about 2 years before the Principia). 1 Figure 13-3(a) is a reproduction of a small fragment of the work, indicating Newton's approach to the problem. With the help of an enlarged version of his sketch [Figure 13-3(b)] we can more readily follow Newton's argument, which as usual was geometrical. Newton imagines an object traveling along AB and then receiving an irripulse directed toward the point S. As a result it now travels along the line BC instead of Bc. Similar impulses carry it to D, E, and F. To make things quantitative, Newton visualizes the displaccmcnt BC as being, in effect, the combination of the displacement Bc, equal to AB, that the object would have undergone if it had continued for an equal length of time with its original velocity, together with the displacement cC parallel to the line BS along which the impulse was applied. This at once yields the Iaw of areas by a simple argument: The triangles SAB and SBc are equal, having equal bases {AB and Bc) and the same altitude. The triangles SBc and SBC are equal, having a common base (SB) and lying between the same parallels. Hence ASAB = ASBC. THE CONSERVATION OF ANGULAR MOMENTUM We give a more modern and more fundamental slant to the law of areas by expressing it in terms of the conservation of orbital angular momentum. If a particle at P [Fig. 13-4(a)] is acted on by a force F, we have & F = /na = m-7- dt Let us now form the vcetor (cross) produet of the position veetor r with both sides of this equation: rXF = rXm^- (13-11) at The left-hand side is the torque M due to F about O. 'This tract, called De Motn (Concerning the Motion of Bodies), contains many of the important results that were later incorporated in the Principia. 560 Motion under central forees Fig. 13-4 (a) Vector relationship ofposi- lion, linear momentum, and orbital momen- tum, (b) Analysis of the velocity into radial and transverse com- ponents. We now introduce the moment of the momentum, I, of the particle with respect to O'. l = rX/wv = rXp (13-12) Thus 1 bears the same relation to the linear momentum, p, as the moment of the force, M, does to F. It is a vector, as shown in Fig. 13-4(a), perpendicular to the plane defined by r and v. If we calculate the rate of change of 1 — i. e., the derivative of 1 with respect to / — we have d\ dt w , w dv Jt = ji x m y+rxm- = \Xmv + rXm dt dt However, the value of v X mv is zero, because it is the cross product of two parallel vectors. The second term on the right of the above equation is, by Eq. (13-1 1), equal to the torque or moment of the applied force about O. Thus we have the simple result M=rXF=^ dt (13-13) This, then, relates the moment of F to the rate of change of the moment of momentum, or orbital angular momentum, of the particle about the origin O. If, now, we take F to be a central force direeted radially toward or away from O, the value of M is zero. In this case, therefore, we have ?=° dt 561 The conservation of angular momentum and 1 = r X p = const. (13-14) This, then, is the statement of the conservation of orbital angular momentum for motion under a central force — or, of course, under zero force. If we look at the situation in the plane of the motion [Fig. 13-4(b)] the vector 1 points upward from the page, and its scalar magnitude / is given by 2 de I = rmo» = mr — dt (13-15a) This means that the constant rate of sweeping out area, as ex- pressed by Eqs. (13-10a) and (13-10b), is given quantitatively by dA _\_ 2 d8 = J_ dt ~ 2 T dt ~ 2m (13-15b) The result expressed by Eqs. (13-14) and (13-15) is a valu- able one. As we have pointed out before, if we find quantities that remain constant throughout some process — i.e., they are "conserved" — these bccome powerful tools in the analysis of phenomena. Angular momentum is a conserved quantity of this sort which is particularly valuable in the analysis of central field problems. It should be noted that the conservation of angular momentum depends only on the absence of a torque and is inde- pendent of the conservation or nonconservation of mechanical energy. A nice example of this last feature is the speeding up of an object that whirls around at the end of a gradually shortening string (Fig. 13-5). If the decrease of r in one revolution is small compared to r itself, the velocity v is almost perpendicular to r at each instant, and the angular momentum (rmo g ) is very nearly equal to mur. The tension T in the string that pulls the object inward exerts no torque about O, and so we have / = const. In a change of r from r\ to r 2 we thus have mv\r\ = mwi Fig. 13-5 Small portion of the path of an object that is being whirled around O at the end of a string the length of which is heing steadily shorlened. 562 Moiion under central forees or r\ V2 = — 01 Thus, if r 2 is less than n, there is a gain of kinetic energy given by K 2 - Kt = JW (^ - l) The work equivalent to this gain of kinetic energy must be pro- vided by the tensi